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ARCHITECTURE. 

BUILDING — CARPENTEY — STAIRS/  ETC. 


THE  ARCHITECT'S  AND  BUILDER'S  POCKET  BOOK 

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THE  THEORY  OP  TRANSVERSE  STRAINS. 

HATFIELD.  And  its  Application  to  the  Construction  of  Buildings,  includ- 
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THE  POETRY  OF  ARCHITECTURE. 

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AN  INQUIRY  INTO  SOME  OF  THE  CONDITIONS 
AFFECTING  "THE  STUDY  OF  ARCHITEC- 
TURE" IN  OUR  SCHOOLS.  By  John  Ruskin. 

12mo,  paper,    0  10 


THEORY 


OF 


TRANSVERSE  STRAINS 


AND    ITS    APPLICATION 


IN   THE 


CONSTRUCTION  OF  BUILDINGS, 


INCLUDING  A  FULL  DISCUSSION  OF  THE  THEORY  AND  CONSTRUCTION  OF  FLOOR 
BEAMS,  GIRIJERS,  HEADERS,  CARRIAGE  BEAMS,  BRIDGING,  ROLLED-IRON  BEAMS, 
TUBULAR  IRON  GIRDERS,  CAST-IRON  GIRDERS,  FRAMED  GIRDERS,  AND  ROOF 
TRUSSES  ;  WITH 

TABLES, 

Calculated  and  prepared  expressly  for  this  Work, 

OF  THE  DIMENSIONS  OF  FLOOR  BEAMS,  HEADERS  AND  ROLLED-IRON  BEAMS  ;  AND 
TABLES  SHOWING  RESULTS  OF  ORIGINAL  EXPERIMENTS  ON  THE  TENSILE,  TRANS- 
VERSE, AND  COMPRESSIVE  STRENGTHS  OF  AMERICAN  WOODS. 

BY 

R.  G.  HATFIELD,  ARCHITECT. 

FELLOW  AM.  INST.  OF  ARCHITECTS  ;  MEM.  AM.  SOC.  OF  CIVIL  ENGINEERS; 
AUTHOR   OF     "AMERICAN    HOUSE   CARPENTER." 


THIRD  EDITION,  REVISED  AND  ENLARGED. 


JOHN   WILEY   &   SONS,    15   ASTOR   PLACE. 

1889. 


COPYRIGHT,  1877. 
JOHN   WILEY  &  SONS. 


PREFACE. 


THIS  work  is  intended  for  architects-  and  students  of  architec- 
ture. 

Within  the  last  ten  years,  many  books  have  been  written  upon 
the  mathematics  of  construction.  Among  them  are  several  of 
particular  excellence.  Few,  however,  are  of  a  character  adapted 
to  the  specific  wants  of  the  architect.  The  subject  is  treated,  by 
some,  in  the  abstract,  and  in  a  manner  so  diffuse  and  'general  as 
to  be  useful  only  to  instructors.  In  other  works,  where  a  prac- 
tical application  is  made,  the  wants  of  the  civil  engineer  rather 
than  of  the  architect  are  consulte.4.  Writers  of  scientific  books, 
as  well  as  the  -public  at  large,  have  failed  to  appreciate  the  wants 
of  the  architect.  Indeed,  many  architects  are  content  to  forego  a 
knowledge  of  construction  ;  following  precedent  as  far  as  pre- 
cedent will  lead,  and,  for  the  rest,  trusting  to  the  chances  of  mere 
guess-work.  For  such,  all  scientific  works  are  alike  useless ;  but 
there  is  a  class  of  architects  who,  through  a  faulty  system  of  edu- 
cation, have  failed  to  obtain,  while  students,  the  knowledge  they 
need  ;  and  who  now  have  little  time  and  less  inclination  to  apply 
themseh^es  to  abstract  or  inappropriate  works,  although  feeling 
keenly  the  need  of  some  knowledge  which  will  help  them  in  their 
daily  duties. 

For  this  class,  and  for  students  in  architecture,  this  book  is 
written.  In  fitting  it  for  its  purpose,  the  course  adopted  has 
been  to  present  an  idea  at  first  in  concrete  form,  and  then  to  lead 
the  mind  gradually  to  the  abstract  truth  or  first  principles  upon 
which  the  idea  is  based.  This  method,  or  the  manner  in  which  it 
is  executed,  may  not  meet  the  approval  of  all.  Nevertheless,  it  is 
hoped  that  those  for  whom  the  work  is  written  may,  by  its  help, 
acquire  the  knowledge  they  need,  and  be  enabled  to  solve  readily 
the  problems  arising  in  their  professional  practice. 


4  PREFACE. 

To  adapt  the  work  to  the  attainments  of  younger  students, 
the  attempt  has  been  made  to  present  the  ideas,  especially  in  the 
first  chapters,  in  a  simple  manner,  elaborating  them  to  a  greater 
extent  than  is  usual. 

The  graphical  method  of  illustration  has  been  employed 
largely,  and  by  its  help  some  of  the  more  abstruse  parts  of  the 
science  of  construction,  it  is  thought,  have  been  made  plain. 
Results  obtained  by  this  method  have  been  analyzed  and  shown 
to  accord  with  the  analytical  formulas  heretofore  employed.  In 
a  discussion  of  the  relation  between  strength  and  stiffness,  a 
method  has  been  developed  for  determining  the  factor  of  safety 
in  the  rules  for  strength.  Rules  for  carriage  beams  with  two  and 
three  headers  are  given.  The  subject  of  bridging  has  been  dis- 
cussed, and  the  value  of  this  system  of  stiffening  floors  defined. 

Especial  attention  has  been  given  to  the  chapters  on  tubular 
iron  girders,  rolled-iron  beams,  framed  girders  and  roofs;  and 
these  chapters,  it  is  hoped,  will  be  particularly  acceptable  to 
architects. 

The  rules  for  the  various  timbers  of  floors,  trussed  girders, 
and  roof  trusses,  are  all  accompanied  by  practical  examples 
worked  out  in  detail.  Tables  are  given  containing  the  dimen- 
sions of  floor  beams  and  headers  for  all  floors.  These  tables  are 
in  two  classes ;  one  for  dwellings  and  assembly  rooms,  the  other 
for  first-class  stores;  and  give  dimensions  for  beams  of  Georgia 
pine,  spruce,  white  pine  and  hemlock,  and  for  rolled-iron  beams. 

Immediately  following  the  tables  will  be  found  a  directory, 
or  digest,  by  which  the  more  important  formulas  are  so  classified 
that  the  proper  one  for  any  particular  use  may  be  discerned  at  a 
glance. 

The  occurrence  recently  of  conflagrations,  resulting  in  serious 
loss  of  life,  has  shown  the  necessity  of  using  every  expedient  cal- 
culated to  render  at  least  our  public  buildings  less  liable  to 
destruction  by  fire.  To  this  end  it  is  proposed  to  construct  timber 
floors  solid,  laying  the  beams  in  contact,  so  as  to  close  the  usual 
spaces  between  the  beams,  and  thus  prevent  the  passage  of  air, 
and  thereby  retard  the  flames.  The  strength  of  these  solid  floors 
has  been  discussed  in  Article  702,  and  a  rule  been  obtained  for  the 
depth  of  beam  or  thickness  of  floor.  By  this  rule  the  depths  for 
floors  of  various  spans  have  been  computed,  and  the  results  re- 
corded in  table  XXI. 


PREFACE.  5 

Tables  XXIII.  to  XLVI.  contain  a  record  of  experiments  made, 
expressly  for  this  work,  upon  six  of  our  American  woods.  In 
these  experiments  and  in  computations,  the  author  has  been  as- 
sisted by  his  son,  Mr.  R.  F.  Hatfield. 

In  the  preparation  of  the  work,  he  has  had  recourse  to  the 
works  of  numerous  writers  on  the  strength  of  materials,  to 
whom  he  is  under  obligation,  and  here  makes  his  acknowledg- 
ments. The  following  are  the  works  which  were  more  particu- 
larly consulted  : — 

Baker  on  Beams,  Columns,  and  Arches. 

Barlow  on  Materials  and  on  Construction. 

Bow  on  Bracing. 

Bow's  Economics  of  Construction. 

Campin  on  Iron  Roofs. 

Cargill's  Strains  upon  Bridge  Girders  and  Roof  Trusses. 

Clark  on  the  Britannia  and   Conway  Tubular  Bridges. 

Emerson's  Principles  of  Mechanics. 

Fairbairn  on  Cast  and  Wrought  Iron. 

Fenwick  on  the  Mechanics  of  Construction. 

Francis  on  the  Strength  of  Cast-Iron  Pillars. 

Haswell's  Engineers'  and  Mechanics'  Pocket-Book. 

Haupt  on  Bridge  Construction. 

Hodgkinson's  Tredgold  on  the  Strength  of  Cast-Iron. 

Humber  on. Strains  in  Girders. 

Hurst's  Tredgold  on  Carpentry. 

Kirkaldy's  Experiments  on  Wrought-Iron  and  Steel. 

Mahan's  Civil  Engineering. 

Mahan's  Moseley's  Engineering  and  Architecture. 

Moseley's  Engineering  and  Architecture. 

Poisson's  Traiie  de  Mecanique. 

Ranken  on  Strains  in  Trusses. 

Rankine's  Applied  Mechanics. 

Robison's  Mechanical  Philosophy. 

Rondelet  sur  le  Dome  du  Pantheon  Fran§ais. 

Sheilds'  Strains  on  Structures  of  Ironwork. 

Styffe  on  Iron  and  Steel. 

Tarn  on  the  Science  of  Building. 

Tate  on  the  Strength  of  Materials. 

Tredgold's  Carpentry. 

Unwin  on  Iron  Bridges  and  Roofs. 

Weisbach's  Mechanics  and  Engineering. 

Wood  on  the  Resistance  of  Materials. 


GENERAL  CONTENTS. 


INTRODUCTION. 


CHAPTER    I. 

THE    LAW    OF    RESISTANCE. 

CHAPTER.    II. 

APPLICATION    OF    THE   LEVER   PRINCIPLE. 

CHAPTER   III. 

DESTRUCTIVE   ENERGY    AND    RESISTANCE. 

CHAPTER   IV. 

EFFECT    OF    WEIGHT    AS    REGARDS   ITS    POSITION. 

CHAPTER   V. 

COMPARISON   OF    CONDITIONS — SAFE   LOAD. 

CHAPTER   VI. 

APPLICATION    OF    RULES — FLOORS. 

CHAPTER  VII. 

GIRDERS,    HEADERS    AND    CARRIAGE   BEAMS. 

CHAPTER   VIII. 

GRAPHICAL     REPRESENTATIONS. 

CHAPTER   IX. 

STRAINS  REPRESENTED  GRAPHICALLY. 

CHAPTER  X. 

STRAINS   FROM   UNIFORMLY    DISTRIBUTED   LOADS. 

CHAPTER   XI. 

STRAINS   IN    LEVERS,   GRAPHICALLY    EXPRESSED. 


GENERAL  CONTENTS. 
CHAPTER   XII. 

COMPOUND    STRAINS    IN    BEAMS,    GRAPHICALLY    EXPRESSED. 

CHAPTER   XIII. 

DEFLECTING    ENERGY. 

CHAPTER  XIV. 

RESISTANCE    TO    FLEXURE. 

CHAPTER    XV. 

RESISTANCE    TO    FLEXURE — LIMIT    OF    ELASTICITY. 

CHAPTER    XVI. 

RESISTANCE    TO    FLEXURE — RULES. 

CHAPTER   XVII. 

RESISTANCE    TO    FLEXURE — FLOOR   BEAMS. 

CHAPTER    XVIII. 

BRIDGING    FLOOR    BEAMS. 

CHAPTER   XIX. 

ROLLED-IRON    BEAMS. 

CHAPTER   XX. 

TUBULAR    IRON    GIRDERS. 

CHAPTER   XXI. 

CAST-IRON    GIRDERS. 

CHAPTER   XXII. 

FRAMED    GIRDERS. 

CHAPTER   XXIII. 

ROOF    TRUSSES. 

CHAPTER   XXIV. 

TABLES. 


DIGEST    OR    DIRECTORY 


INDEX. 


ANSWERS    TO     QUESTIONS. 


CONTENTS. 


INTRODUCTION. 

ART. 

1. — Construction  Defined 

2. — Stability  Indispensable. 

3. — Laws  Governing  the  Force  of  Gravity. 

4. — Science  of  Construction,  for  Architect  rather  than  Builder. 

5. — Parts  of  Buildings  requiring  Special  Attention. 

6. — This  Work  Limited  to  the  Transverse  Strain. 

7. — In  Construction  Safety  Indispensable. 

8. — Some  Floors  are  Deficient  in  Strength. 

9. — Precedents  not  always  Accessible. 
10. — An  Experimental  Floor,  an  Expensive  Test. 
11. — Requisite  Knowledge  through  Specimen  Tests. 
12. — Unit  of  Material — Its  Dimensions. 
13. — Value  of  the  Unit  for  Seven  Kinds  of  Material. 


CHAPTER    I. 

THE    LAW   OF    RESISTANCE. 

14. — Relation  between  Size  and  Strength. 

15. — Strength  not  always  in  Proportion  to  Area  of  Cross-section. 

16. — Resistance  in  Proportion  to  Area  of  Cross-section. 

17. — Units  may  be  Taken  of  any  Given  Dimensions. 

18. — Experience  Shows  a  Beam  Stronger  when  Set  on  Edge. 

19. — Strength  Directly  in  Proportion  to  Breadth. 

20.— By  Experiment  Strength  Increases  more  Rapidly  than  the  Depth. 

2  1. — Comparison  of  a  Solid  Beam  with  a  Laminated  one. 

22. — Strength  due  to  Resistance  of  Fibres  to  Extension  and  Compression. 

23, — Power  Extending  Fibres  in  Proportion  to  Depth  of  Beam. 


CHAPTER    II. 

APPLICATION    OF   THE    LEVER    PRINCIPLE. 

24.— The  Law  of  the  Lever. 

25. — Equilibrium — Direction  of  Pressures. 


CONTENTS. 

260 — Conditions  of  Pressure  in  a  Loaded  Beam. 

27. — The  Principle  of  the  Lever. 

28. — A  Loaded  Beam  Supported  at  Each  End. 

29.— A  Bent  Lever. 

30. — Horizontal  Strains  Illustrated  by  the  Bent  Lever. 

31. — Resistance  of  Fibres  in  Proportion  to  the  Depth  of  Beam. 


CHAPTER   III. 

DESTRUCTIVE    ENERGY   AND   RESISTANCE. 

32. —  Resistance  to  Compression — Neutral  Line. 

33. — Elements  of  Resistance  to  Rupture. 

34. — Destructive  Energies. 

35.— Rule  for  Transverse  Strength  of  Beams. 

36. — Formulas  Derived  from  this  Rule. 

37  to  51. — Questions  for  Practice. 


CHAPTER   IV. 

THE   EFFECT   OF   WEIGHT   AS   REGARDS    ITS    POSITION. 

52. — Relation  between  Destructive  Energy  and  Resistance. 
53. — Dimensions  and  Weights  to  be  of  Like  Denominations  with  those  of 
the  Unit  Adopted. 

54. — Position  of  the  Weight  upon  the  Beam. 

55. — Formula  Modified  to  Apply  to  a  Lever. 

56. — Effect  of  a  Load  at  Any  Point  in  a  Beam. 

57.— Rule  for  a  Beam  Loaded  at  Any  Point. 

58.— Effect  of  an  Equally  Distributed  Load. 

59.— Effect  at  Middle  from  an  Equally  Distributed  Load. 

60.— Example  of  Effect  of  an  Equally  Distributed  Load. 

61. — Result  also  Obtained  by  the  Lever  Principle. 

62  to  65. — Questions  for  Practice. 


CHAPTER    V. 

COMPARISON   OF   CONDITIONS — SAFE    LOAD. 

66. — Relation  between  Lengths,  Weights  and  Effects. 
67.— Equal  Effects. 


10  CONTENTS. 

68. — Comparison  of  Lengths  and  Weights  Producing  Equal   Effects. 

69.  —  Tne  Effects  from  Equal  Weights  and  Lengths. 

70. — Rules  for  Gases  in  which  the  Weights  and  Lengths  are  Equal. 

71. — Breaking  and  Safe  Loads. 

72. — The  above  Rules  Useful  Only  in  Experiments. 

73 — Value  of    a,     the  Symbol  of  Safety. 

74.  — Value  of    a,     the  Symbol  of  Safety. 

75 — Rules  for  Safe  Loads. 

76. — Applications  of  the  Rules. 

77. — Example  of  Load  at  End  of  Lever. 

78. — Arithmetical  Exemplification  of  the  Rule. 

79.— Caution  in  Regard  to     a,     the  Symbol  of  Safety. 

80. — Various  Methods  of  Solving  a  Problem. 

81. — Example  of  Uniformly  Distributed  Load  on  Lever. 

82. — Load  Concentrated  at  Middle  of  Beam. 

83. — Load  Uniformly  Distributed  on  Beam  Supported  at  Both  Ends. 

84  to  87. — Questions  for  Practice. 


CHAPTER   VI. 

APPLICATION    OF    RULES — FLOORS. 

88. — Application  of  Rules  to  Construction  of  Floors. 

89. — Proper  Rule  for  Floors. 

90. — The  Load  on  Ordinary  Floors,  Equally  Distributed. 

9  I . — Floors  of  Warehouses,  Factories  and  Mills. 

92. — Rule  for  Load  upon  a  Fioor  Beam. 

93. — Nature  of  the  Load  upon  a  Floor  Beam. 

94. — Weight  of  Wooden  Beams. 

95. — Weight  in  Stores,  Factories  and  Mills  to  be  Estimated. 

96.— Weight  of  Floor  Plank. 

97. — Weight  of  Plastering. 

98. — Weight  of  Beams  in  Dwellings. 

99. — Weight  of  Floors  in  Dwellings. 
100. — Superimposed  Load. 
101. — Greatest  Load  upon  a  Floor. 
102.— Tredgold's  Estimate  of  Weight  on  a  Floor. 
103. — Tredgold's  Estimate  not  Substantiated  by  Proof. 
104. — Weight  of  People — Sundry  Authorities. 
105. — Estimated  Weight  of  People  per  Square  Foot  of  Floor. 
106. — Weight  of  People,  Estimated  as  a  Live  Load. 
107.— Weight  of  Military. 

103. — Actual  Weights  of  Men  at  Jackson's  and  at  Hoes'  Foundries. 
109. — Actual  Measure  of  Live  Load. 

II  0.— More  Space  Required  for  Live  Load. 

III  . — No  Addition  to  Strain  by  Live  Load. 


CONTENTS.  1 1 

1 1  2. — Margin  of  Safety  Ample  for  Momentary  Extra  Strain  in  Extreme  Cases. 
1 1  3. — Weight  Reduced  by  Furniture  Reducing  Standing  Room. 
1 1 4.— The  Greatest  Load  to  be  provided   for  is    70    Pounds  per   Super- 
ficial Foot. 

1 1  5. — Rule  for  Floors  of  Dwellings. 

116. — Distinguishing  Between  Known  and  Unknown  Quantities. 

117' — Practical  Example. 

118. — Eliminating  Unknown  Quantities. 

1 1  9. — Isolating  the  Required  Unknown  Quantity. 

120. — Distance  from  Centres  at  Given  Breadth  and  Depth. 

1  21. — Distance  from  Centres  at  Another  Breadth  and  Depth. 

1  22. — Distance  from  Centres  at  a  Third  Breadth  and  Depth. 

123. — Breadth,  the  Depth  and  Distance  from  Centres  being  Given. 

124.  — Depth,  the  Breadth  and  Distance  from  Centres  being  Given. 

125. — General  Rules  for  Strength  of  Beams. 

126  to  135. — Questions  for  Practice. 


CHAPTER   VII. 

GIRDERS,    HEADERS   AND   CARRIAGE    BEAMS. 

136 — A  Girder  Denned. 

137 — Rule  for  Girders. 

138.— Distance  between  Centres  of  Girders. 

139.— Example  of  Distance  from  Centres. 

140. — Size  of  Girder  Required  in  above  Example. 

141. — Framing  for  Fireplaces,  Stairs  and  Light-wells. 

142.— Definition  of  Carriage  Beams,  Headers  and  Tail  Beams. 

143. — Formula  for  Headers — General  Considerations. 

144.— Allowance  for  Damage  by  Mortising. 

145.— Rule  for  Headers. 

146. — Example.  • 

147.— Carriage  Beams  and  Bridle  Irons. 

148,-Rule  for  Bridle  Irons. 

149.— Example. 

150. — Rule  for  Carriage  Beam  with  One  Header. 

151. — Example. 

152.— Carriage  Beam  with  Two  Headers. 

153. — Effect  of  Two  Weights  at  the  Location  of  One  of  Them. 

154.— Example. 

155. — Rule  for  Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams. 

156. — Example. 

157. — Rule  for  Carriage  Beam  with  Two  Headers  and  One  Set  of  Tail  Beams. 

158.— Example. 

159  to  166.— Questions  for  Practice. 


12  CONTENTS. 

CHAPTER   VIII. 

GRAPHICAL   REPRESENTATIONS. 


167. — Advantages  of  Graphical  Representations. 

168. — Strains  in  a  Lever  Measured  by  Scale. 

169.— Example — Rule  for  Dimensions. 

170. — Graphical  Strains  in  a  Double  Lever. 

171.— Graphical  Strains  in  a  Beam. 

172.— Nature  of  the  Shearing  Strain. 

173. — Transverse  and  Shearing  Strains  Compared. 

174.— Rule  for  Shearing  Strain  at  Ends  of  Beams. 

175. — Resistance  to  Side  Pressure. 

176. — Bearing  Surface  of  Beams  upon  Walls. 

177. — Example  to  Find  Bearing  Surface. 

178.— Shape  of  Side  of  Beam,  Graphically  Expressed. 

179  to  187. — Questions  for  Practice. 


CHAPTER    IX. 

STRAINS  REPRESENTED  GRAPHICALLY. 


188. — Graphic  Method  Extended  to  Other  Cases. 

189. — Application  to  Double  Lever  with  Unequal  Arms. 

190.— Applicati6n  to  Beam  with  Weight  at  Any  Point. 

191.— Example. 

192.— Graphical  Strains  by  Two  Weights. 

193. — Demonstration. 

194. — Demonstration— Rule  for  the  Varying  Depths. 

195.— Graphical  Strains  by  Three  Weights. 

196.— Graphical  Strains  by  Three  Equal  Weights  Equably  Disposed. 

197.— Graphical  Strains  by  Four  Equal  Weights  Equably  Disposed. 

198.^Graphical  Strains  by  Five  Equal  Weights  Equably  Disposed. 

199._General  Results  from  Equal  Weights  Equably  Disposed. 

200.— General  Expression  for  Full  Strain  at  First  Weight. 

201.— General  Expression  for  Full  Strain  at  Second  Weight. 

202.— General  Expression  for  Full  Strain  at  Any  Weight. 

203. — Example. 

204  to  209. — Questions  for  Practics. 


CONTENTS.  1 3 

CHAPTER   X. 

STRAINS    FROM    UNIFORMLY    DISTRIBUTED    LOADS. 

210. — Distinction    Between   a    Series    of    Concentrated    Weights    and    a 
Thoroughly  Distributed  Load. 
211 1 — Demonstration. 
212. — Demonstration  by  the  Calculus. 
213. — Distinction  Shown  by  Scales  of  Strains. 
214. — Effect  at  Any  Point  by  an  Equally  Distributed  Load. 
215. — Shape  of  Side  of  Beam  for  an  Equably  Distributed  Load. 
216. — The  Form  of  Side  of  Beam  a  Semi-ellipse. 
217  to  220. — Questions  for  Practice. 


CHAPTER   XL 

STRAINS    IN    LEVERS,    GRAPHICALLY    EXPRESSED. 

221. — Scale  of  Strains  for  Promiscuously  Loaded  Lever. 

222. — Strains  and  Sizes  of  Lever  Uniformly  Loaded. 

223. — The  Form  of  Side  of  Lever  a  Triangle. 

224. — Combinations  of  Conditions. 

225. — Strains  and  Dimensions  for  Compound  Load. 

226. — Scale  of  Strains  for  Compound  Loads. 

227. — Scale  of  Strains  for  Promiscuous  Load. 

228  to  233. — Questions  for  Practice. 


CHAPTER   XII. 

COMPOUND   STRAINS    IN    BEAMS,    GRAPHICALLY    EXPRESSED. 

234. — Equably  Distributed  and  Concentrated  Loads  on  a  Beam. 

235. — Greatest  Strain  Graphically  Represented. 

236. — Location  of  Greatest  Strain  Analytically  Defined. 

237. — Location  of  Greatest  Strain  Differentially  Defined. 

238. — Greatest  Strain  Analytically  Defined. 

239. — Example. 

240. — Dimensions  of  Beam  for  Distributed  and  Concentrated  Loads. 

241. — Comparison  of  Formulas,  Here  and  in  Art.  150. 

242. — Location  of  Greatest  Strain  Differentially  Defined, 

243. — Greatest  Strain  and  Dimensions. 

244. — Assigning  the  Symbols. 

245. — Example — Strain  and  Size  at  a  Given  Point. 

246.— Example— Greatest  Strain. 

247. — Example — Dimensions. 


14  CONTENTS. 

248, — Dimensions  for  Greatest  Strain  when     h     Equals     n. 

249. — Dimensions  for  Greatest  Strain  when     //     is  Greater  than     n. 

250i — Rule  for  Carriage  Beams  with  Two  Headers  and  Two  Sets  of  Tail 
Beams. 

251. — Example. 

2.52. — Carriage  Beam  with  Three  Headers 

253,— Three  Headers — Strains  of  the  First  Class. 

254. — Graphical  Representation. 

255. — Greatest  Strain. 

256.— General  Rule  for  Equably  Distributed  and  Three  Concentrated  Loads. 

257.— Example. 

258. — Rule  for  Carriage  Beams  with  Three  Headers  and  Two  Sets  of  Tail 
Beams. 

259.— Example. 

260.— Three  Headers— Strains  of  the  Second  Class. 

261. — Greatest  Strain. 

262. — General  Rule  for  Equally  Distributed  and  Three  Concentrated  Loads. 

263. — Example. 

264. — Assigning  the  Symbols. 

265. — Reassigning  the  Symbols. 

266.— Example. 

267.— Rule  for  Carriage  Beam  with  Three  Headers  and  Two  Sets  of  Tail 
Beams. 

268.— Example. 

269  and  270.— Questions  for  Practice. 


CHAPTER   XIII. 

DEFLECTING     ENERGY. 

271.— Previously  Given  Rules  are  for  Rupture. 

272.— Beam  not  only  to  Be  Safe,  but  to  Appear  Safe. 

273. — All  Materials  Possess  Elasticity. 

274.— Limits  of  Elasticity  Denned. 

275.— A  Knowledge  of  the  Limits  of  Elasticity  Requisite. 

276.— Extension  Directly  as  the  Force. 

277.— Extension  Directly  as  the  Length. 

278, — Amount  of  Deflection. 

279.— The  First  Step. 

280. —Deflection  to  be  Obtained  from  the  Extension. 

281.— Deflection  Directly  as  the  Extension. 

282. — Deflection  Directly  as  the  Force,  and  as  the  Length. 

283. —  Deflection  Directly  as  the  Length. 

284.— Deflection  Directly  as  the  Length. 

285. — Total  Deflection  Directly  as  the  Cube  of  the  Length. 

286. — Deflecting  Energy  Directly  as  the  Weight  and  Cube  of  the  Length. 

287  to  291.— Questions  for  Practice. 


CONTENTS.  1 5 

i 
CHAPTER   XIV. 

RESISTANCE   TO  FLEXURE. 

292, — Resistance  to  Rupture,  Directly  as  the  Square  of  the  Depth. 
293. — Resistance  to  Extension  Graphically  Shown. 

294. — Resistance  to  Extension  in  Proportion  to  the  Number  of  Fibres  and 
their  Distance  from  Neutral  Line. 
295. — Illustration. 

296. — Summing  up  the  Resistances  of  the  Fibres. 
297. — True  Value  to  which  these  Results  Approximate. 
298.— True  Value  Denned  by  the  Calculus. 

299. — Sum  of  the  Two  Resistances,  to  Extension  and  to  Compression. 
300. — Formula  for  Deflection  in  Levers. 
301. — Formica,  for  Deflection  in  Beams. 
302. — Value  of    F,   the  Symbol  for  Resistance  to  Flexure. 
303.— Comparison  of    F    with    £,    the  Modulus  of  Elasticity. 
304.— Relative  Value  of    F    and     E. 

305. — Comparison  of    F    with    E    common,  and  with  the    E    of  Barlow. 
306. — Example  under  the  Rule  for  Flexure. 
307  to  310.— Questions  for  Practice. 


CHAPTER   XV. 

RESISTANCE  TO   FLEXURE— LIMIT   OF   ELASTICITY. 

311. — Rules  for  Rupture  and  for  Flexure  Compared. 
312.— The  Value  of    a,     the  Symbol  for  Safe  Weight. 
313. — Rate  of  Deflection  per  Foot  Length  of  Beam. 
314. — Rate  of  Deflection  in  Floors. 
315  to  319.— Questions  for  Practice. 


CHAPTER  XVI. 

RESISTANCE  TO   FLEXURE — RULES. 

320. — Deflection  of  a  Beam,  with  Example. 
321. — Precautions  as  to  Values  of  Constants     F    and     <?. 
322.— Values    of    Constants    F    and    e     to    be    Derived    from    Actual 
Experiment  in  Certain  Cases. 


1 6  CONTENTS. 

323. — Deflection  of  a  Lever. 

324.— Example. 

325.— Test  by  Rule  for  Elastic  Limit  in  a  Lever. 

326. —  Load  Producing  a  Given  Deflection  in  a  Beam. 

327.— Example. 

328. — LoaJ  at  the  Limit  of  Elasticity  in  a  Beam. 

329, — Load  Producing  a  Given  Deflection  in  a  Lever — Example 

330.--Deflection  in  a  Lever  at  the  Limit  of  Elasticity. 

331. — Load  on  Lever  at  the  Limit  of  Elasticity. 

332.— Values  of     W,    /,    b,   d    and    c5     in  a  Beam. 

333. — Example — Value  of    /    in  a  Beam. 

334. — Example — Value  of    b     in  a  Beam. 

335. — Example — Value  of    d    in  a  Beam. 

336.— Values  of    P,    n,    l>,   d    and     6    in  a  Lever. 

337. — Example — Value  of    «     in  a  Lever. 

338. — Example— Value  of    b     in  a  Lever. 
339.— Example— Value  of    d    in  a  Lever. 
340. — Deflection — Uniformly  Distributed  Load  on  a  Beam. 
341. — Values  of     U,    I,    b,    d    and     <5     in  a  Beam, 
342. — Example — Value  of     U,     the  Weight,  in  a  Beam. 
343. — Example — Value  of     /,     the  Length,  in  a  Beam. 
344. — Example — Value  of    l>,     the  Breadth,  in  a  Beam. 
345. — Example — Value^of    d,     the  Depth,  in  a  Beam. 
346. — Example — Value  of    (5,     the  Deflection,  in  a  Beam. 
347. — Deflection — Uniformly  Distributed  Load  on  a  Lever. 
348.— Values  of     U,    n,   (>,    d    and     d     in  a  Lever. 
349.— Example— Value  of     U,     the  Weight,  in  a  Lever. 
350.— Example — Value  of    «,     the  Length,  in  a  Lever. 
351. — Example — Value  of    b,     the  Breadth,  in  a  Lever, 
352. — Example — Value  of    </,     the  Depth,  in   a  Lever, 
353. — Example — Value  of    (5,     the  Deflection,  in  a  Lever. 
354  to  357. — Questions  for  Practice. 


CHAPTER  XVII. 

RESISTANCE  TO    FLEXURE— FLOOR    BEAMS. 

358. — Stiffness  a  Requisite  in  Floor  Beams. 

359. — General  Rule  for  Floor  Beams. 

360.— The  Rule  Modified. 

361.— Rule  for  Dwellings  and  Assembly  Rooms. 

362. — Rules  giving  the  Values  of    c,    /.    b    and     &. 

363.— Example— Distance  from  Centres, 

364.— Example— Length, 

365. — Example — Breadth. 

366. — Example — Depth. 


CONTENTS.  I 

367. — Floor  Beams  for  Stores. 
368. — Floor  Beams  of  First-class  Stores. 
369. — Rule  for  Beams  of  First-class  Stores. 
370.— Values  of    c,    I,    b     and     d. 
371. — Example — Distance  from  Centres. 
372. — Example — Length. 
373.— Example — Breadth. 
374.— Example— Depth. 
375. — Headers  and  Trimmers. 

376. — Strength  and  Stiffness — Relation  of  Formulas. 
377. — Strength  and  Stiffness — Value  of    a,     in  Terms  of    B    and     F. 
378.— Example. 
379.— Test  of  the  Rule. 

380. — Rules  for  Strength  and  Stiffness  Resolvable. 
381.— Rule  for  the  Breadth  of  a  Header. 
382. — Example  of  a  Header  for  a  Dwelling. 
383. — Example  of  a  Header  in  a  First  class  Store. 
384, — Carriage  Beam  with  One  Header. 
385. — Carriage  Beam  with  One  Header,  for  Dwellings. 
386. — Example. 

387. —Carriage  Beam  with  One  Header,  for  First-class  Stores. 
388.— Example. 

389. — Carriage  Beam  with  One  Header,  for  Dwellings — More  Precise  Rule. 
390.— Example. 

391. — Carriage  Beam  with  One  Header,  for  First-class  Stores — More  Pre- 
cise Rule. 

392.— Example. 

393. — Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams,  for 
Dwellings,  etc. 

394.— Example. 

395. — Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams,  for 
Firsl-clasj  Stores. 
396.— Example. 

397. — Carriage  Beam  with  Two  Headers  and  One  Set  of  Tail  Beams. 
398, — Carriage  Beam  with  Two   Headers  and   One  Set  of  Tail  Beams,  for 
Dwellings. 

399.— Example. 

400.— Carriage  Beam  with  Two   Headers  and  One   Set  of  Tail  Beams,  for 
First-class  Stores. 
401. — Example. 

402.— Carriage  Beam  with  Two  Headers  and   Two  Sets  of  Tail  Beams — 
More  Precise  Rules. 

403.— Example—  h     less  than     ». 
404.— Example—  //     greater  than     n. 

405. — Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams,  for 
Dwellings — More  Precise  Rule. 
406. — Example. 

407.— Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams,  for 
First-class  Stores — More  Precise  Rule. 


1 8  CONTENTS. 

408.— Example. 

409.— Carriage   Beam  with  Two  Headers  and  One  Set  of  Tail  Beams — 
More  Precise  Rule. 

410.— Example. 

411. — Carriage  Beam  with  Two  Headers  and  One  Set  of  Tail   Beams,  for 
Dwellings— More  Precise  Rule. 

412. — Example. 

413. — Carriage  Beam  with  Two  Headers  and  One  Set  of  Tail   Beams,  for 
First-class  Stores — More  Precise  Rule. 

414.— Example. 

415. — Carriage  Beam  with  Two    Headers,  Equidistant    from    Centre,  and 
Two  Sets  of  Tail  Beams— Precise  Rule. 

416. — Example. 

417i — Carriage  Beams  with  Two  Headers,  Equidistant  from  Centre,  and  Two 
sets  of  Tail  Beams,  for  Dwellings  and  for  First-class  Stores— Precise  Rules. 

418. — Examples 

419. — Carriage  Beam  with  Two  Headers,  Equidistant  from  Centre,  and  One 
Set  of  Tail  Beams— Precise  Rule. 

420.— Example. 

421. — Carriage  Beams  with  Two   Headers,  Equidistant  from  Centre,    and 
One  Set  of  Tail  Beams,  for  Dwellings  and  for  First-class  Stores — Precise  Rules. 

422.— Example. 

423. —Beam  with  Uniformly  Distributed  and  Three  Concentrated  Loads, 
the  Greatest  Strain  being  Outside. 

424.— Example. 

425. — Carriage  Beam  with  Three  Headers,  the  Greatest   Strain  being  at 
Outside  Header. 

426. — Example. 

427. — Carriage   Beam   with   Three   Headers,  the  Greatest  Strain  being  at 
Outside  Header,  for  Dwellings. 

428. — Carriage  Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Outside  Header,  for  First-class  Stores. 

429.— Examples. 

430. — Beams  with  Uniformly  Distributed  and  Three  Concentrated  Loads, 
the  Greatest  Strain  being  at  Middle  Load. 

431. — Example. 

432. — Carriage   Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Middle  Header. 

433.— Example. 

434. — Carriage  Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Middle  Header,  for  Dwellings. 

435. — Carriage  Beam   with  Three  Headers,  the  Greatest  Strain  being  at 
Middle  Header,  for  First-class  Stores. 

436.— Example. 

437  to  442. — Questions  for  Practice. 


CONTENTS. 
CHAPTER   XVIII. 

BRIDGING    FLOOR   BEAMS. 

443,— Bridging  Defined. 

444. — Experimental  Test. 

445. — Bridging — Principles  of  Resistance. 

446. — Resistance  of  a  Bridged  Beam 

447.— Summing  the  Resistances. 

448.— Example. 

449. — Assistance  Derived  from  Cross-bridging. 

450. — Number  of  Beams  Affording  Assistance. 

451. — Bridging  Useful  in  Sustaining  Concentrated  Weights. 

452. — Increased  Resistance  Due  to  Bridging. 


CHAPTER   XIX. 

ROLLED-IRON     BEAMS. 

453.— Iron  a  Substitute  for  Wood. 
454. — Iron  Beam — Its  Progressive  Development. 
455. — Rolled-Iron  Beam — Its  Introduction. 
456. — Proportions  between  Flanges  and  Web. 
457. — The  Moment  of  Inertia  Arithmetically  Considered. 
458.— Example  A. 
459.— Example  B. 
460.— Example  C. 
461. — Comparison  of  Results. 

462. — Moment  of  Inertia,  by  the  Calculus — Preliminary  Statement. 
463. — Moment  of  Inertia,  by  the  Calculus. 
464. — Application  and  Comparison. 
465.— Moment  of  Inertia  Graphically  Represented. 
466. — Parabolic  Curve — Area  of  Figure. 
467. — Example. 

468. — Moment  of  Inertia — General  Rule. 
469. — Application. 

470.— Rolled-Iron  Beam — Moment  of  Inertia — Top  Flange. 
471. — Rolled-Iron  Beam — Moment  of  Inertia — Web. 
472. — Rolled-Iron  Beam — Moment  of  Inertia— Flange  and  Web. 
473. — Rolled-Iron  Beam— Moment  of  Inertia — Whole   Section. 
474. — Rolled-Iron   Beam— Moment   of    Inertia — Comparison   v/ith    other 
Formulas. 

475. — Rolled-Iron  Beam — Moment  of  Inertia — Comparison  of  Results. 
476. — Rolled-Iron  Beam — Moment  of  Inertia — Remarks. 
477.— Reduction  of  Formula — Load  at  Middle. 


2O  CONTENTS. 

478.— Rules— Values  of     W,    /,     <5    and    7. 

479.— Example— Weight. 

480. — Example— Length. 

481.— Example— Deflection. 

482. — Example — Moment  of  Inertia. 

483. — Load  at  Any  Point— General  Rule. 

484.— Load  at  Any  Point  on  Rolled-Iron  Beams. 

485. — Load  at  Any  Point  on  Rolled-Iron  Beams  of  Table  XVII. 

486. — Example. 

487. — Load  at  End  of  Rolled-Iron  Lever. 

488.— Example. 

489.— Uniformly  Distributed  Load  on  Rolled-Iron  Beam. 

490.— Example. 

491. — Uniformly  Distributed  Load  on  Rolled-Iron  Lever. 

492.— Example. 

493.— Components  of  Load  on  Floor. 

494. — The  Superincumbent  Load. 

495.— The  Materials  of  Construction— Their  Weight. 

496.— The  Rolled-Iron  Beam— Its  Weight. 

497.— Total  Load  on  Floors. 

498. — Floor  Beams — Distance  from  Centres. 

499.— Example. 

500.— Floor  Beams — Distance  from  Centres — Dwellings,  etc. 

501. — Example. 

502. — Floor  Beams — Distance  from-Centres. 

503. — Example. 

504.— Floor  Beams — Distance  from  Centres— First-class  Stores. 

505.  -Example. 

506. — Floor  Arches — General  Considerations. 

507.— Floor  Arches— Tie  Rods. 

508.— Example. 

509.— Headers. 

510. — Headers  for  Dwellings,  etc. 

511. — Example. 

512.— Headers  for  First-class  Stores. 

513. — Carriage  Beam  with  One  Header. 

514. — Carriage  Beam  with  One  Header,  for  Dwellings,  etc. 

515. — Example. 

516. — Carriage  Beam  with  One  Header,  for  First-class  Stores. 

517. — Example. 

518. — Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams. 

519. — Caniage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams,  for 
Dwellings,  etc 

520.— Example. 

521. — Carriage  Beam  with  Two  Headers  and  Two  Sets  of  Tail  Beams,  for 
First-class  Stores. 

522.— Example. 

523. — Carriage  Beam  with  Two  Headers,  Equidistant  from  Centre,  and 
Two  Sets  of  Tail  Beams,  for  Dwellings,  etc. 


CONTENTS.  21 

524. — Example. 

525. — Carriage  Beam  with  Two  Headers,  Equidistant  from  Centre,  and 
Two  Sets  of  Tail  Beams,  for  First-class  Stores. 

526.— Example. 

527. — Carriage  Beam  with  Two  Headers  and  One  Set  of  Tail  Beams,  for 
Dwellings,  etc. 

528.— Example. 

529.  —Carriage  Beam  with  Two  Headers  anal  One  Set  of  Tail  Beams,  for 
First-class  Stores. 

530.— Example. 

531. — Carriage  Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Outside  Header,  for  Dwellings,  etc. 

532. — Example. 

533. — Carriage  Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Outside  Header,  for  First-class  Stores. 

534. — Carriage  Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Middle  Header,  for  Dwellings,  etc. 

535. — Example. 

536. — Carriage  Beam  with  Three  Headers,  the  Greatest  Strain  being  at 
Middle  Header,  for  First-class  Stores. 

537  to  545.— Questions  for  Practice. 


CHAPTER    XX. 

TUBULAR   IRON    GIRDERS. 

546. — Introduction  of  the  Tubular  Girder. 

547. — Load  at  Middle — Rule  Essentially  the  Same  as  that  for  Rolled-Iron 
Beams. 

548. — Load  at  Any  Point — Load  Uniformly  Distributed. 

549. — Load  at  Middle — Common  Rule. 

550.— Capacity  by  the  Principle  of  Moments. 

551.— Load  at  Middle — Moments. 

552. — Example. 

553. — Load  at  Any  Point. 

554. — Example. 

555.— Load  Uniformly  Distributed. 

556. — Example. 

557. — Thickness  of  Flanges. 

558. — Construction  of  Flanges. 

559. — Shearing  Strain. 

560.— Thickness  of  Web. 

561. — Example. 

562.— Construction  of  Web. 

563.— Floor  Girder— Area  of  Flange. 

564.— Weight  of  the  Girder. 


22  CONTENTS. 

565. — Weight  of  Girder  per  Foot  Superficial  of  Floor. 

566. — Example. 

567.— Total  Weight  of  Floor  per  Foot  Superficial,   including  Girder. 

568. — Girders  for  Floors  of  Dwellings,  etc. 

569. — Example. 

570. — Girders  for  Floors  of  First-class  Stores. 

571. — Ratio  of  Depth  to  Length,  in  Iron  Girders. 

572. — Economical  Depth. 

573. — Example. 

574  to  579.— Questions  for  Practice. 


CHAPTER   XXI. 

CAST-IRON     GIRDERS. 

580. — Cast  Iron  Superseded  by  Wrought  Iron. 

581. — Flanges — Their  Relative  Proportion. 

582.— Flanges  and  Web — Relative  Proportion. 

583.— Load  at  Middle. 

584.— Example. 

585.— Load  Uniformly  Distributed. 

586.— Load  at  Any  Point— Rupture. 

587. — Safe  Load  at  Any  Point. 

588.— Example. 

589.— Safe  Load  Uniformly  Distributed— Effect  at  Any  Point. 

590.— Form  of  Web. 

591.— Two  Concentrated  Weights — Safe  Load. 

592.— Examples. 

593.— A  relied  Girder. 

594.— Tie-Rod  of  Arched  Girder. 

595.— Example. 

596. — Substitute  for  Arched  Girder. 

597  to  602. — Questions  for  Practice. 


CHAPTER   XXII. 

FRAMED    GIRDERS. 

603.^-Transverse  Strains  in  Framed  Girders. 
604.— Device  for  Increasing  the  Strength  of  a  Beam. 
605. — Horizontal  Thrust. 

606. — Parallelogram  of  Forces — Triangle  of  Forces. 
607. — Lines  and  Forces  in  Proportion. 


CONTENTS.  23 

608, — Horizontal  Strain  Measured  Graphically. 

609, — Measure  of  Any  Number  of  Forces  in  Equilibrium. 

610. — Strains  in  an  Equilibrated  Truss. 

611,— From  Given  Weights  to  Construct  a  Scale  of  Strains. 

612, — Example. 

613, — Horizontal  Strain  Measured  Arithmetically. 

614. — Vertical  Pressure  upon  the  Two  Points  of  Support. 

615. — Strains  Measured  Arithmetically. 

616. — Curve  of  Equilibrium — Stable  and  Unstable. 

617. — Trussing  a  Frame. 

618. — Forces  in  a  Truss  Graphically  Measured. 

619. — Example. 

620. — Another  Example. 

621.— Diagram  of  Forces. 

622. — Diagram  of  Forces — Order  of  Development. 

623. — Reciprocal  Figures. 

624. — Proportions'in  a  Framed  Girder. 

625. — Example. 

626.— Trussing,  in  a  Framed  Girder. 

627. — Planning  a  Framed  Girder. 

628.— Example. 

629. — Example. 

630. — Number  of  Bays  in  a  Framed  Girder. 

631.— Forces  in  a  Framed  Girder. 

632.  — Diagram  for  the  above  Framed  Girder. 

633.— Gradation  of  Strains  in  Chords  and  Diagonals. 

634. — Framed  Girder  with  Loads  on  Each  Chord. 

635.— Gradation  of  Strains  in  Chords  and  Diagonals. 

636.— Strains  Measured  Arithmetically. 

637. — Strains  in  the  Diagonals. 

638. — Example. 

639.— Strains  in  the  Lower  Chord. 

640.— Strains  in  the  Upper  Chord. 

641. — Example. 

642.— Resistance  to  Tension. 

643. — Resistance  to  Compression. 

644. — Top  Chord  and  Diagonals— Dimensions. 

645. — Example. 

646. — Derangement  from  Shrinkage  of  Timbers. 

647. — Framed  Girder  with  Unequal  Loads,  Irregularly  Placed. 

648. — Load  upon  Each  Support — Graphical  Representation. 

649. — Girder  Irregularly  Loaded — Force  Diagram. 

650. — Load  upon  Each  Support,  Arithmetically  Obtained. 

651  to  656, — Questions  for  Practice. 


24  CONTENTS. 

CHAPTER   XXIII. 

ROOF   TRUSSES. 

657. — Roof  Trusses  considered  as  Framed  Girders. 

658. — Comparison  of  Roof  Trusses. 

659. — Force  Diagram — Load  upon  Each  Support. 

660. — Force  Diagram  for  Truss  in  Fig.  98. 

661. — Force  Diagram  for  Truss  in  Fig.  99. 

662. — Force  Diagram  for  Truss  in  Fig.  zoo- 

663. — Force  Diagram  for  Truss  in  Fig.  101. 

664. — Force  Diagram  for  Truss  in  Fig.  102. 

665. — Force  Diagram  for  Truss  in  Fig.  103. 

666. — Force  Diagram  for  Truss  in  Fig.  104. 

667. — Force  Diagram  for  Truss  in  Fig.  105. 

668. — Force  Diagram  for  Truss  in  Fig.  106. 

669. — Strains  in  Horizontal  and  Inclined  Ties  Compared. 

670.— Vertical  Strain  in  Truss  with  Inclined  Tie. 

671.— Illustrations. 

672.— Planning  a  Roof. 

673.— Load  upon  Roof  Truss. 

674.— Load  on  Roof  per  Foot  Horizontal. 

675. — Load  upon  Tie-Beam. 

676. — Selection  of  Design  for  Roof  Truss. 

677. — Load  on  Each  Supported  Point  in  Truss. 

678. — Load  on  Each  Supported  Point  in  Tie-Beam. 

679. — Constructing  the  Force  Diagram. 

680.— Measuring  the  Force  Diagram. 

681. — Strains  Computed  Arithmetically. 

682. — Dimensions  of  Parts  Subject  to  Tension. 

683.— Dimensions  of  Parts  Subject  to  Compression. 

684.— Dimensions  of  Mid-Rafter. 

685. — Dimensions  of  Upper  Rafter. 

686. — Dimensions  of  Brace. 

687. — Dimensions  of  Straining-Beam. 

688  to  692.— Questions  for  Practice. 


CHAPTER   XXIV. 

TABLES. 

693.— Tables  I.  to  XXL— Their  Utility. 

694 — Floor  Beams  of  Wood  and  Iron  (I.  to  XIX.). 

695 — Floor  Beams  of  Wood  (I.  to  VIII.). 

696 — Headers  of  Wood  (IX.  to  XVI.). 

697 — Elements  of  Rolled-Iron  Beams  (XVII.). 

638.— Rolled-Iron  Beams  for  Office  Buildings,  etc.  (XVIIL). 


CONTENTS.  25 

699.  -Rolled-Iron  Beams  for  First-class  Stores  (XIX.). 

700. — Example. 

701. — Constants  for  Use  in  the  Rules  (XX.). 

702. -Solid  Timber  Floors  (XXI.). 

703.— Weights  of  Building  Materials  (XXII.). 

704.— Experiments  on  American  Woods  (XXIII.  to  XLVI.). 

705.— Experiments  by  Transverse  Strain  (XXIII.  to  XXXV.,  XLU.  and 
XLIII.). 

706.— Experiments  by  Tensile  and  Sliding  Strains  (XXXVI.  to  XXXIX., 
XLIV.  and  XLV.). 

707.— Experiments  by  Crushing  Strain  (XL.,  XLI.  and  XLVI.). 


TABLES. 

I. — Hemlock  Floor  Beams  for  Dwellings,  Office  Buildings,  etc. 
II.— White  Pine 
III.— Spruce 
IV. — Georgia  Pine     " 

V. — Hemlock  Floor  beams  for  First-class  Stores. 
VI.— White  Pine 
VII.— Spruce 
VIII.— Georgia  Pine    " 

IX. — Hemlock  Headers  for  Dwellings,  Office  Buildings,  etc. 
X.— White  Pine       " 
XI.— Spruce 
XII.— Georgia  Pine    " 

XIII. — Hemlock  Headers  for  First-class  Stores. 
XIV.— White  Pine       " 
XV.— Spruce 

XVI.— Georgia  Pine  "         "  " 

XVII.— Elements  of  Rolled-Iron  Beams. 

XVIII.— Rolled  Iron  Beams  for  Dwellings,  Office  Buildings,  etc. 
XIX. —  "  "         "    First-class  Stores. 

XX. — Values  of  Constants  Used  in  the  Rules. 
XXL— Solid  Timber  Floors— Thickness. 
XXII.— Weights  of  Materials  of  Construction  and  Loading. 
XXIII. — Transverse  Strains  in  Georgia  Pine. 
XXIV.—          "  "       "    Locust. 

XXV.—          '•  "       "   White  Oak. 

XXVI.—          "  "       "    Spruce. 

XXVII.—          "  

XXVIIL—          "  "       " 

XXIX.—          "  "       "   White  Pine. 

XXX.—          "  "       " 

XXXI.—          "  "       " 

XXXII.—          "  "       " 


26  CONTENTS. 

XXX III.— Transverse  Strains  in  Hemlock. 
XXXIV.—  "  "       " 

XXXV.—          "  "       " 

XXXVI.— Tensile  Strains  in  Georgia  Pine,  Locust  and  White  Oak. 
XXXVII.—      "  "       "  Spruce,  White  Pine  and  Hemlock. 

XXXVIII.— Sliding  Strains  in  Georgia  Pine,  Locust  and  White  Oak. 
XXXIX.—      "  "       "  Spruce,  White  Pine  and  Hemlock. 

XL. — Crushing  Strains  in  Georgia  Pine,  Locust  and  White  Oak. 
XLI. —      "  "       "  Spruce,  White  Pine  and  Hemlock. 

XLII. — Rupture  by  Transverse  Strain — Values  of    B. 
XLIII. — Resistance  to  Deflection — Values  of    F, 
XLIV. — Rupture  by  Tensile  Strain — Values  of    7". 
XLV.—        "         "    Sliding       "  "       "     G. 

XLVI. —        "         "   Compressive  Strain — Values  of    C. 


DIGEST   OR   DIRECTORY. 

INDEX. 
ANSWERS  TO   QUESTIONS. 


INTRODUCTION. 


ART.  I. — The  science  of  Construction,  as  the  term  is  used 
in  architecture,  comprehends  a  knowledge  of  the  forces 
tending  to  destroy  the  materials  constituting  a  building,  and 
of  the  capacities  of  resistance  of  the  materials  to  these  forces. 

2. — One  of  the  requisites  of  good  architecture  is  Sta- 
bility. Without  this  the  beautiful  designs  of  the  architect 
can  have  no  lasting  existence  beyond  the  paper  upon  which 
they  are  delineated. 

3. — The  force  of  Gravity  is  inherent  not  only  in  the 
contents  of  a  building,  but  also  in  the  materials  of  which  the 
building  itself  is  constructed ;  and  unless  these  materials 
have  an  adequate  power  of  resistance  to  this  force,  the  safety 
of  the  building  is  endangered.  Hence  the  necessity  of  a 
knowledge  of  the  laws  governing  the  force  of  gravity  in  its 
action  upon  the  several  parts  of  a  building,  and  of  the  expe- 
dients to  be  resorted  to  in  order  to  resist  its  action  effect- 
ually. 

4-. — It  may  be  objected  by  some  that  this  knowledge 
pertains  rather  to  building  than  to  architecture,  and  that 
the  architect  is  required  merely  to  indicate  the  outlines 
of  his  plans,  leaving  to  the  builder  the  work  of  deter- 
mining the  arrangement  and  dimensions  of  the  materials. 
This  objection  is  not  well  founded.  Between  the  duties  of 


28  INTRODUCTION. 

the  architect  and  those  of  the  builder  there  is  a  well-defined 
line.  This  may  be  shown  by  a  consideration  of  the  operation 
of  building  as  it  is  usually  conducted.  The  builder  is  selected 
generally  from  among  those  who  compete  for  the  work. 
Each  builder  competing  fixes  the  amount  for  which  he  is 
willing  to  erect  the  building,  after  an  examination  of  the  plans 
and  specifications  and  an  estimate  of  the  cost  of  the  work. 
To  arrive  at  this  cost  the  arrangement  and  dimensions  of  the 
materials  must  be  fixed  ;  and  if  not  fixed  by  the  plans  and 
specifications,  in  what  way  shall  they  be  determined  ?  Shall 
it  be  by  the  builder  ?  The  builder  has  not  yet  been  selected. 
Shall  each  builder  estimating  be  permitted  to  assign  such  di- 
mensions as  his  caprice  or  cupidity  shall  dictate  ?  The  evil 
effect  of  such  a  course  is  apparent.  The  only  proper  method 
is  to  have  the  arrangement  and  dimensions  of  the  materials 
all  definitely  settled  by  the  architect  in  his  plans  and  specifi- 
cations. 

Moreover,  the  necessity  for  a  knowledge  of  this  subject 
by  the  architect  is  manifest  in  this,  that  he  is  constantly  liable, 
without  this  knowledge,  to  include  in  his  plans  such  features 
as  the  action  of  gravity  would  render  impossible  of  produc- 
tion in  solid  material,  or  which,  if  executed,  would  not  pos- 
sess the  requisite  degree  of  stability. 

5. — In  considering  the  requisites  for  stability  in  a  build- 
ing, the  various  parts  need  to  be  taken  in  detail :  such  as 
Walls,  Piers,  Columns,  Buttresses,  Foundations,  Arches, 
Lintels,  Floors,  Partitions,  Posts,  Girders  and  Roofs. 

6. — It  is  the  purpose  of  the  present  work  to  treat 
principally  of  those  parts  which  are  subjected  to  trans- 
verse strains. 

7. — In  the  construction  of  a  floor,  the  safety  of  those 
who  are  to  trust  themselves  upon  it  is  the  first  consideration. 


TO   OBTAIN   A   RULE   FOR   FLOOR   TIMBERS.  29 

8. — Floors  are  not  always  made  sufficiently  strong. 
Scarcely  a  year  passes  without  its  record  of  deaths  conse- 
quent upon  the  failure  of  floors  upon  which  people  should 
have  assembled  with  safety.  Many  floors  now  existing, 
and  not  a  few  of  those  annually  constructed,  are  deficient 
in  material,  or  have  an  improper  arrangement  of  it. 

9. — The  strength  of  a  floor  consists  in  the  strength  of  its 
timbers. 

The  dimensions  of  the  timbers  for  any  given  floor  may  be 
ascertained,  practically,  by  an  examination  of  other  similar 
floors  which  have  been  tried  and  found  sufficiently  strong. 
But  if  no  similar  floor  is  found,  how  is  the  problem  to  be 
solved  ? 

10. — The  amount  of  material  required  may  be  found 
by  constructing  one  or  more  experimental  floors,  and  testing 
them  with  proper  weights ;  but  this  wrould  be  attended 
with  great  expense,  and  probably  with  the  loss  of  more  time 
than  could  be  spared  for  the  purpose. 

II. — There  is  a  simple  method,  which  is- quite  ascertain 
and  less  expensive.  The  chemist,  from  a  small  specimen, 
makes  an  analysis  sufficient  to  determine  the  character  of 
whole  mines  of  ore  or  quarries  of  rock.  So  we,  by  proper 
tests  of  a  small  piece  of  any  building  material,  may  deter- 
mine the  characteristics  of  ail  material  of  that  kind. 

12. — To  obtain,  then,  the  requisite  knowledge  of  the 
strength  of  floor  timbers,  let  us  adopt  a  piece  of  convenient 
size  as  the  unit  of  material.  Let  it  be  a  piece  one  inch  square 
and  one  foot  long  in  the  clear  between  the  bearings.  This 
we  will  submit  to  a  transverse  force,  applied  at  the  middle  of 
its  length,  sufficient  to  break  it  crosswise,  and  learn  from 
the  result  the  power  of  resistance  it  possesses. 

Numerous  experiments  of  this  nature  have   been    made 


30  INTRODUCTION. 

upon  all  the  ordinary  kinds  of  timber,  stone  and  iron,  and 
the  average  results  collected  in  tabular  form.  (See  Table 
XX.)  A  few  results  are  here  given. 

13. — The  unit  of  material,  when  of 

Hemlock,      breaks  with  450  pounds : 
White  Pine,       "          "     500 
Spruce,  "          "     550 

White  Oak,  "  "  650 
Georgia  Pine,  "  "  850 
Locust,  "  "  1200  " 

Cast-Iron  "          "  2100         " 

These  figures  give  the  average  unit  of  strength  for  these 
several  kinds  of  material,  when  exposed  to  a  transverse 
strain  at  the  middle  of  their  length. 


CHAPTER  I. 

THE   LAW   OF   RESISTANCE. 

ART.  14. — Relation  between  Size  and  Strength. — Having 
ascertained,  by  careful  experiment,  the  power  of  resistance 
in  a  unit  of  any  given  material,  the  next  question  is :  What 
is  the  existing  relation  between  size  and  strength  ?  Is  the 
increase  of  one  proportionate  to  that  of  the  other  ? 

In  two  square  beams  of  equal  length,  but  of  different 
sectional  area,  the  larger  one  will  bear  more  than  the 
smaller.  From  this  it  appears  that  the  resistance  is,  to  a 
certain  degree  at  least,  in  proportion  to  the  quantity  of 
material,  or  to  the  area  of  cross-section.  There  is  an 
element  of  strength,  however,  other  than  this,  and  one  which 
modifies  the  proportion  very  materially. 

IS. — Strength  not  aBway§  in  Proportion  to  Area  of 
Cross-§eeiioii. — That  the  strength  of  any  two  pieces  of  equal 
length  is  not  always  in  proportion  to  the  area  of  cross-sec- 
tion, is  shown  by  attempts  to  break  two  given  pieces.  For 
example,  take  two  beams  of  equal  length,  but  of  differing 
area  of  cross-section;  the  one  being  3  x  8,  and  the  other 
5x6  inches.  The  former  has  24  and  the  latter  30  inches  of 
sectional  area.  If  the  strength  be  in  proportion  to  the 
sectional  area,  the  weights  required  to  break  these  two 
pieces  will  be  in  the  proportion  of  24  to  30 — their  relative 
areas  of  cross-section ;  but  they  will  be  found  (the  pieces 
being  placed  upon  edge)  to  be  in  the  proportion  of  24  to 
the  smaller  piece  being  actually  stronger  than  the  larger ! 


32  THE   LAW   OF   RESISTANCE.  CHAP.    I. 


16.  —  Resistance  in  Proportion  to  Area  of 

—  Preliminary  to  seeking  the  cause  of  this  apparent  want 
of  proportion,  it  will  be  well  to  show  first  that,  under  certain 
conditions,  the  resistance  of  beams  is  directly  proportional 
to  their  area  of  cross-section. 

Let  there  be  twenty  pieces  of  smooth  white  pine,  each 
one  inch  square,  and  one  foot  long  in  the  clear  between  the 
bearings.  The  resistance  of  anyone  of  these  pieces  is  limited 
to  500  pounds.  This  has  been  ascertained  by  experiment  as 
before  stated  in  Art.  13. 

Let  four  of  these  pieces  be  placed  side  by  side  upon  the 
bearings.  The  resistance  of  the  four  is  evidently  just  four 
times  the  resistance  of  one  piece  ;  or  4  x  500  =  2000  pounds. 

Let  four  more  pieces  be  placed  upon  the  first  four  :  the 
strength  of  the  eight  amounts  to  2  x  2000  =  4000  pounds. 

Add  four  more,  and  the  combined  resistance  of  the  twelve 
pieces  will  be  3  x  2000  =  6000  pounds. 

The  resistance  of  four  tiers  of  four  each,  or  of  sixteen 
pieces,  will  be  16  x  500  =  Soco  pounds. 

The  total  strength  of  the  twenty  pieces,  piled  up  five  tiers 
high,  will  be  20  x  500  =10,000  pounds. 

Thus  we  see  that  the  resistance  is  exactly  in  proportion 
to  the  amount  of  material  used.* 

17.  —  Units  may  be  Taken   of  any  Given    Dimensions.  — 

In  this  trial  we  have  taken  as  the  unit  of  material  a  bar  one 
inch  square.  Wve  might  have  taken  this  unit  of  any  other 
dimension,  as  a  half,  a  quarter,  or  even  a  tenth  of  an  inch 
square,  and,  after  finding  by  trial  the  strength  of  one  of 


*  The  truth  of  this  proposition  depends  upon  obtaining,  in  the  experiment, 
pieces  of  wood  so  smooth  that,  in  being  deflected  by  the  weight,  they  will  move 
upon  each  other  without  friction  ;  a  condition  not  quite  possible  in  practice  to 
obtain.  This  friction  restrains  free  action,  and,  as  a  consequence,  the  weight 
required  to  effect  the  rupture  will  be  somewhat  greater  than  is  stated. 


RELATION  BETWEEN  AREA  AND  STRENGTH.       33 

these  units,  could  have  as  readily  known  the  strength  of  the 
whole  pile  by  merely  multiplying  the  number  of  units  by 
the  strength  of  one  of  them. 

We  will  now  consider  the  relation  between  breadth  and 
depth. 


18. — Experience  Snows  a  Beam  Stronger  when  Set  on 
Edge. — One  of  the  first  lessons  of  experience  with  timber 
of  greater  breadth  than  thickness,  is  the  fact  of  its  pos- 
sessing greater  strength  when  placed  on  edge  than  when 
laid  on  the  flat.  As  an  example  :  a  beam  of  white  pine, 
3x8  inches,  and  10  feet  long  between  bearings,  will 
require  9600  pounds  to  break  it  when  set  on  edge  ;  while 
three  eighths  of  this  amount,  or  3600  pounds,  will  break  it  if 
it  be  laid  upon  the  flat.  Here  again,  as  in  Art.  15,  we  have 
a  fact  seemingly  at  variance  with  the  one  but  just  previously 
established — namely,  that  of  the  resistance  being  in  propor- 
tion to  the  area  of  cross-section.  We  will  now  investigate 
the  apparent  anomaly. 


(9. — Strength  IMrectly  in  Proportion  to  Breadth. — First, 
as  to  the  breadth  of  a  beam.  If  two  beams  of  like  size  are 
placed  side  by  side,  the  two  will  resist  just  twice  that  which 
one  of  them  alone  would.  Three  beams  will  resist  three 
times  as  much  as  one  beam  would.  So  of  any  number  of 
beams,  the  resistance  will  be  in  proportion  directly  as  the 
breadth. 

This  is  found  by  trial  to  be  true,  whether  the  beams  are 
separate  or  together,  solid ;  for  a  6  x  8  inch  beam  will  bear 
as  much,  and  only  as  much,  as  three  beams  2x8  inches  set 
side  by  side,  and,  in  both  cases,  on  the  edge.  In  other 
words,  when  the  depths  and  lengths  are  equal,  a  beam  of  six 


34  THE    LAW   OF   RESISTANCE.  CHAP.    L 

inches  breadth  will  bear  just  three  times  as  much  as  a  beam 
of  two  inches  breadth,  or  twice  as  much  as  one  of  three 
inches  breadth. 

So  this  fact  appears  established,  that  the  resistance  of 
beams  is  directly  in  proportion  to  their  breadth. 

20. — By  Experiment  Strength  Iiicrcasc§  more  Rapidly 
than  the  Depth. — In  regarding  the  depth  of  beams,  another 
law  of  proportion  is  found.  Having  two  beams  of  the  same 
breadth  and  length,  but  differing  in  depth,  we  find  the 
strength  greater  than  in  proportion  to  the  depth.  If  it  were 
in  this  proportion,  a  beam  nine  inches  high  would  bear 
just  three  times  as  much  as  one  three  inches  high,  whereas 
experiment  shows  it  to  bear  much  more  than  this. 

21. — Comparison  of  a  Solid  Beam  with  Jt  Laminated 
one. — To  test  this,  let  there  be  two  beams  of  equal  length, 
breadth  and  depth,  one  of  them  being  in  one  solid  piece 


FIG.  i.  FIG.  2. 

(Fig.  2),  while  the  other  is  made  up  of  horizontal  layers 
or  veneers,  laid  together  loosely  (Fig.  i).  Placing  weights 
upon  these  two  beams,  it  is  seen  that,  although  they  contain 
a  like  quantity  of  material  in  cross-section,  and  are  of  equal 
height,  the  solid  beam  will  sustain  much  more  weight  than 
the  laminated  one.  Let  the  several  parts  of  the  latter  beam 
be  connected  together  by  glue,  or  other  cementing  material, 


CHANGE   IN   LENGTH   OF   FIBRES.  35 

and  again  applying  weights,  it  will  be  found  that  it  has  be- 
come nearly,  if  not  quite,  as  strong  as  the  beam  naturally 
solid. 

From  these  results  we  infer  that  the  increased  strength  is 
due  to  the  union  of  the  fibres  at  each  juncture  of  the  hori- 
zontal layers.  But  why  does  this  result  follow?  If  the 
simple  knitting  together  of  the  fibres  is  the  cause,  then  why, 
in  considering  the  breadth,  is  a  solid  beam  no  stronger 
than  two  beams,  each  of  half  the  breadth,  as  has  been 
shown  ? 


22. — Strength  clue  to  Resistance  of  Fibres  to  Extension 
and  Compression. — An  examination  of  the  action  of  the 
beams  under  pressure  in  Figs,  i  and  2  may  explain  this. 
The  weights  bending  the  beams  make  them  concave  on  top. 
In  Fig.  i  the  ends  of  the  veneers  or  layers  remain  in  vertical 
planes,  while,  in  the  other  case,  the  end  of  the  solid  beam 
is  inclined,  and  normal  to  the  curve.  It  is  also  seen  that 
the  upper  surface  in  Fig.  2  is  shorter  than  the  lower  one, 
although  the  two  surfaces  were  of  the  same  length  before 
bending.  This  change  in  length  has  occurred  during  the 
process  of  bending,  and  could  only  happen  through  a  change 
in  length  of  the  fibres  constituting  the  beam. 

In  the  operation  of  bending,  one  of  two  things  must  of 
necessity  take  place  :  either  the  fibres  must  slide  upon  each 
other,  as  in  Fig.  i,  or  else  the  length  of  the  fibres  must  be 
changed,  as  in  Fig.  2  ;  and  since  in  practice  it  is  found  that 
the  fibres  are  so  firmly  knit  as  effectually  to  prevent  sliding, 
we  have  only  to  consider  the  effects  of  a  change  in  the 
length  of  the  fibres.  The  resistance  to  this  change  is  an 
element  of  strength  other  than  that  due  to  quantity  of 
material,  and  its  nature  will  now  be  examined. 


3^  THE   LAW   OF   RESISTANCE.  CHAP.   I. 

23. — Power  Extending  Fibres  in  Proportion  to  Depth 
of  Beam. — If  a  beam  be  made  of  four  equal  pieces,  as  in 
Fig-  3>  and  be  held  together  by  an  elastic  strap  firmly 

attached  to  the  under  side 
of  the  beam,  and  by  two 
cross  pieces  let  into  the 
horizontal  joint  and  closely 
fitted  ;  and  if  upon  this  beam 
FlG-  3-  a  weight  be  laid  at  the 

middle  sufficient  to  elongate  the  strap  and  open  the  vertical 
joint  at  the  bottom  a  given  distance — say  an  eighth  of  an 
inch ;  then,  if  the  weight  and  the  two  upper  quarters  of  the 
beam  be  removed,  and  a  weight  laid  on  at  the  middle  suffi- 
cient to  open  the  joint,  to  the  like  distance  as  before,  it  will 
be  found  that  this  weight  is  just  one  half  of  that  before  used. 
In  this  experiment,  the  strap  may  be  taken  to  represent  the 
fibres  at  the  lower  edge  of  the  beam. 

We  here  find  a  relation  between  the  weight  and  the 
height  of  the  beam.  The  greater  the  height,  the  greater 
must  be  the  weight  to  produce  a  like  effect  upon  the  fibres 
of  the  lower  edge.  Double  the  height  requires  double  the 
weight.  Three  times  the  height  requires  three  times  the 
weight.  Therefore  we  decide  that,  in  elongating-  the  fibres 
at  the  bottom,  the  weight  and  the  height  are  directly  in  pro- 
portion. 

It  must  be  observed  that  Fig.  3  and  its  explanation  arc 
not  to  be  taken  as  a  representation  of  the  full  effect  of  a 
transverse  strain  upon  a  beam.  The  scope  of  the  experi- 
ment is  limited  to  the  action  of  the  fibres  at  the  lower 
edge.  The  other  fibres,  all  contributing  more  or  less  to  the 
resistance,  are,  for  the  moment,  neglected,  in  order  to  show 
this  one  feature  of  the  strain — namely,  the  manner  in  which 
fibres  at  any  point  contribute  to  the  general  resistance. 

Galileo,  of  Italy,  who,  two  hundred  and  fifty  years  since, 


NEUTRAL   LINE.  37 

was  the  first  to  show  the  connection  of  the  theory  of  trans- 
verse strains  with  mathematics,  not  recognizing  in  his  theory 
the  compressibility  of  the  fibres  at  the  concave  side  of  the 
beam,  supposed  that  in  a  rupture  by  cross  strain  all  the 
fibres  were  separated  by  pulling  apart ;  as  might  be  shown 
in  Fig.  3,  in  case  the  rubber  were  extended  up  each  side  to 
the  top,  instead  of  being  confined  to  the  lower  edge.  We 
are  greatly  indebted  to  Galileo  for  his  studies  in  this  direc- 
tion ;  but  Hooke,  Mariotte,  and  Leibnitz,  about  1680,  found 
the  theory  of  Galileo  to  be  defective,  and  showed  that  the 
fibres  were  elastic ;  that  only  those  fibres  at  the  convex  side 
of  the  beam  suffer  extension ;  that  those  at  the  concave  side 
suffer  compression  and  are  shortened  ;  and  that,  at  the  line 
separating  the  fibres  which  are  extended  from  those  which 
are  compressed,  they  are  neither  lengthened  nor  shortened, 
but  remain  at  their  natural  length.  This  line  is  denominated 
the  neutral  line  or  surface. 

It  will  here  be  observed  that  the  amount  of  extension  or 
compression  in  any  fibre  is  proportional  to  its  distance  from 
the  neutral  line. 


CHAPTER    II. 

APPLICATION   OF   THE   LEVER   PRINCIPLE. 

ART.  24. — The  Law  of  the  L,ever. — The  deduction  drawn 
from  the  experiment  named  in  Art.  23  depends  for  its  truth 
upon  what  is  known  as  the  law  of  the  lever.  This  law,  in 
so  far  as  it  applies  to  transverse  strains,  will  now  be  con- 
sidered. 

25. — Equilibrium— Direction  of  Pre§sures. — When  equal 
weights,  suspended  from  the  ends  of  a  beam  supported  upon 
a  fulcrum,  as  at  W  in  Fig.  4,  are  in  equilibrium,  it  is  found 
that  the  point  of  support 'is  just  midway  between  the  two 
weights,  provided  that  the  beam  be  of  equal  size  and  weight 
throughout  its  length. 

It  will  be  observed  that  the    directions    of  the   strains 
upon  the  beam  are  vertical,  those  at  the  ends  being  down- 
•  w  ward,  while  that  at  the   middle  is 

upward  ;  also  that  the  strains  are 
evidently  equal,  the  upward  pres- 
sure at  the  middle  being  just  equal 
to  the  sum  of  the  two  weights  at 
FIG.  4.  the  ends;  for  if  unequal,  there 

would  be  no  equilibrium,  but  a  movement  in  the  direction 
of  the  greater  power. 

We  decide,  then,  that  the  pressure  upon  the  fulcrum  is 
equal  to  the  sum  of  the  two  weights.* 

*  In  ascertaining  the  pressure  at  the  fulcrum,  the  weight  of  the  beam 
itself  should  be  added  to  the  sum  of  the  two  weights,  but  to  simplify  the  ques- 
tion, the  beam,  or  lever,  is  supposed  to  be  without  weight. 


REACTION   EQUAL  TO   THE   PRESSURE.  39 

25. — Conditions    of  Fre§§ure    in   a   Loaded    Beam. — In 

Fig.  5  we  have  a  beam  supported  at  each  end,  and  a  weight 

W  laid  upon  the  middle  of  its 

length. 

Comparing  this  with  Fig.  4 

we  see  that  the   strains  here    ..j^ ^J^ 

are  also  vertical  but  in  re- 
versed order,  the  one  at  the 
middle  being  downwards,  FIG.  5. 

while  those  at  the  ends  are  upwards.  In  other  respects  we 
have  here  the  same  conditions  as  in  Fig.  4. 

The  downward  pressure  at  the  middle  is  equal  to  the 
upward  pressures  or  reactions  at  the  ends ;  and,  since  the 
weight  is  placed  midway  between  the  points  of  support,  the 
reactions  at  these  points  are  equal,  and  each  is  equal  to  one 
half  the  weight  at  the  middle. 

27. — The  Principle  of  the  Lever. — In  Fig.  6  is  shown  a 
lever  resting  upon  a  fulcrum  Wt  and  carrying  at  its  ends 
the  weights  R  and  P.  w  

Here,  the  fulcrum  W  is  not  at 
the  middle  as  in  Fig.  4,  but  at  a 
point  which  divides  the  lever  into 
two  unequal  parts,  m  and  n. 

In    accordance  with  the  prin-  FIG.  6. 

ciple  of  the  lever,  the  two  parts  m  and  n,  when  there  is  an 
equilibrium,  are  in  proportion  to  the  two  weights  P  and 
R;  or,  the  shorter  arm  is  to  the  longer  as  the  lesser  weight 
is  to  the  greater  ;*  or, 

m  :  n  : :  P  :  R 


*  For  a  demonstration  of  the  lever  principle  see  an  article,  by  the  author, 
in  the  Mathematical  Monthly,  published  at  Cambridge,  U.  S.f  vol.  I,  1858, 
page  77. 


4O  APPLICATION   OF   THE    LEVER   PRINCIPLE.        CHAP.   II. 

from  which  we  have 


Rm  =  Pn 
~  n 


(1.) 
m  v    ' 


and. 

^m 


As  an  example:  suppose  the  lever  to  be  12  feet  long,  and 
so  placed  upon  the  fulcrum  as  to  make  the  two  arms,  m  and 
«,  4  and  8  feet  respectively.  Then,  if  the  shorter  arm  have 
suspended  from  its  end  a  weight,  R,  of  500  pounds,  what 
weight,  P,  will  be  required  at  the  end  of  the  longer  arm  to 
produce  equilibrium  ? 

Formula   (#.)  is    appropriate     to   this    case.      Therefore 

P=R™  —  500  x  ~=  250  pounds;  equals  the  weight  required 

on  the  longer  arm. 

From  Art.  25  it  is  evident  that  the  sum  of  the  weights 
R  and  P  is  equal  to  the  upward  force  or  reaction  at  W. 

Therefore,  we  have, 

W=£  +  P 
and  W-R=P 

Substituting  this  value  for  />in  formula  (^.),we  have 


n 

_  m 

n 

W  =  R  ' 

and,  multiplying  by 


PRESSURE   ON   SUPPORTS     MEASURED.  41 

and,  since  n  +  m  is  equal  to  the  whole  length  of  the  beam, 
or  to  /,  therefore 

R=W^  (3.) 

In  a  similar  manner,  it  is  found  that 

P=  w™  (4.) 

28- — A    Loaded    Beam    Supported   at    Eaeli    End. — In 

Fig.  7  a  weight  W,  is  carried  by  a  beam  resting  at  its  ends 
upon  two  supports.  Here  we 
have,  with  the  pressures  in 
reversed  order,  similar  con- 
ditions with  those  shown  in 
Fig.  6.  Here,  also,  it  will  be 
observed  that  the  weight  W 
is  equal  to  the  sum  of  the 

upward  resistances  R  and  P  (Arts.  25  and  26) — neglecting 
for  the  present  the  weight  of  the  beam  itself— and  that  the 
upward  resistance  at  R  may  be  found  by  formula  (3.) ;  while 
that  at  P  is  found  by  formula  (4.). 

For  example :  suppose  the  weight  W,  Fig.  7,  to  be  800 
pounds ;  and  that  it  be  located  five  feet  from  one  end  of  the 
beam  and  eight  feet  from  the  other  end. 
Here  W  —  800,  m  —  5,  n  =  8  and  /  =  13. 

To  find  the  pressure  at  R,  we  have,  by  formula  (#.), 

R  —  W  —  =  800  x  —   =  4Q2T4T  pounds. 
I  13 

To  find  the  pressure  at  P,  we  have,  by  formula  (4.\ 
P—  W~  —  800  x  --  =  307^  pounds. 

To  verify  the  rule,  we  find  that 

+  3°7A  =  800  pounds  =  W. 


APPLICATION   OF    THE   LEVER   PRINCIPLE.     CHAP.  II. 


Either  one  of  these  upward  pressures,  or  reactions,  being- 
found,  the  other  may  be  determined  by  subtracting1  the  first 
from  W. 

From  the  above,  we  see  that  the  portion  of  a  weight 
borne  by  one  support  is  equal  to  the  product  of  the  weight 
into  its  distance  from  the  other  support,  divided  by  the 
length  between  the  two  supports. 

29.  —  A  Bent  L.evcr.  —  In  Fig.  8  let  P  C  G  be  a  rigid  bar, 

shaped  to  a  right  angle  at  C,  and 
free  to  revolve  on  C  as  a  centre. 
Let  R  and  H  be  two  weights  at- 
tached by  cords  to  the  points  P 
and  G,  the  cords  passing  over 
the  pulleys  D  and  E.  Let  the 
weights  be  so  proportioned  as 
FIG.  8.  to  produce  an  equilibrium. 

Here  P  C  G  is  what  is  termed  a  bent  lever,  and  the 
arms  a  and  b  are  in  proportion  to  the  weights  R  and  H  ;  or, 

a  \  b  \\  R  \  H  and 


30.—  Horizontal  Strains  Illustrated  by  the  Bent  Lever.— 

To  apply  the  principle   of  the  bent  lever  let   a  beam  R  E 

(Fig.  9)  be  laid  upon  two  points  of  support,  R  and  E,  and 

be  loaded  at  the  middle  with 
the  weight  W.  The  action 
-of  this  weight  upon  the  beam 
is  similar  in  its  effect  to  that 
taking  place  in  the  bent  lever 
of  Fig.  8,  producing  horizon- 
FIG.  9.  tal  strains,  which  compress 

the  fibres  at  the  top  of  the  beam  and  extend  those  at  the 

bottom.     (Art.  23). 


HORIZONTAL   STRAIN     MEASURED.  43 

Let  the  line  P  C  represent  the  line  of  division  between 
the  compressed  and  the  extended  fibres.  Then  P  C  G  may 
be  taken  to  represent  the  bent  lever  of  Fig.  8 ;  for  the 
upward  pressure  or  reaction  at  R,  moving  the  arm  of  lever 
P,C,  which  turns  on  the  point  (7,  as  a  centre,  acts  upon  the 
point  G,  through  the  arm  of  lever  *C  G,  moving  the  point  G 
horizontally  from  E,  and  thus  extending  the  fibres  in  the 
line  G  E. 

Now,  if  H  represents  this  horizontal  strain  along  the 
bottom  of  the  beam,  and  -R  the  vertical  strain  at  P—  both 
being  due  to  the  action  of  the  weight  W;  if  the  arm  P  C 
be  called  b,  and  C  G  called  a,  then,  as  before, 

a  :  b  :  :  R  :  H  from  which 

H=Rt 

a 

For  an  application:  let  b  in  a  given  case  equal  10  feet,  a 
equal  6  inches,  or  0-5  of  a  foot,  and  R  equal  1200  pounds; 
what  will  be  the  horizontal  strain  in  the  fibres  at  the  lower 
edge  of  the  beam  ? 

From  the  above  formula, 

„  b  10 

H  --  R—  =  1200  x  —  —  1200x20.=  24,000 

or  the  horizontal  strain  equals  24,000  pounds. 

31. — Re§i§taiicc  of  Fibres  in  Proportion  to  the  Depth  of 

Beam.— From  the  proportion  in  the  last  article, 

a  :  b  : :  R  :  H  we  have 

Ha  =  Rb  and  dividing  by  a  b  we  have 

J^-^-jK 
b         a 

For  any  given  material,  the  power  of  the  fibres  to  resist 
tension  is  limited,  and,  since  this  power  is  represented  by  H, 


44  APPLICATION   OF  THE  LEVER  PRINCIPLE.       CHAP.   II. 

therefore  H  is  limited.  In  any  given  length  of  beam,  b, 
which  is  dependent  upon  the  length,  is  also  given  ;  hence 

TT  TT  -D  r> 

-T-  becomes  a  fixed*  quantity  ;  and  since  -7-  =  — ,  therefore  — 

is  a  fixed  quantity.  But  R  and  a  may  vary  individually, 
provided  that  the  quotient  of  R  divided  by  a  be  not  changed. 
So,  then,  if  R  be  increased,  a  must  also  be  increased,  and  in 
a  like  proportion ;  if  R  be  doubled,  a  must  be  doubled ; 
if  one  be  trebled,  the  other  must  be  trebled  ;  or,  in  whatever 
proportion  one  is  increased  or  diminished,  the  other  must 
be  increased  or  diminished  in  like  proportion.  Therefore 
R  and  a  are  in  direct  proportion. 

Take    a   as    equal    to    one    half   of   the    depth    of    the 

beam,  or  — ,  and  R  as  equal  to  one  half  the  weight  at  the 

W 
middle  of  the  beam,  or  — . 

Then,  since  a  is  in  proportion  to  R,  d  is  in  proportion  to 
W,  or  the  depth  of  the  beam  must  be  in  proportion  to 
the  weight. 

This  result  is  the  same  as  that  arrived  at  in  Art.  23 ; 
that  the  power  of  the  fibres  at  the  bottom  to  resist  extension 
is  in  proportion  to  the  depth  of  the  beam. 


CHAPTER   III. 

DESTRUCTIVE   ENERGY   AND   RESISTANCE. 

ART.  32. — Re§istance  to  Comprc§§ion— Neutral  Line. — We 

have  shown  the  manner  in  which  the  fibres  at  the  convex 
side  of  a  beam  contribute  to  its  strength  by  their  resistance 
to  extension.  It  may  now  be  observed  that  the  resistance  to 
compression  of  the  fibres  at  the  concave  side  is  but  a  counter- 
part of  the  resistance  to  extension  of  the  fibres  at  the  convex 
side. 

Whatever  resistance  may  be  given  out  in  one  way  at  one 
side  of  the  beam,  a  like  amount  of  resistance  will  be  called 
up  in  the  other  way  at  the  other  side.  The  one  balances  the 
other,  like  two  weights  at  the  ends  of  a  lever  (Figs.  4  and  6). 
If  the  powers  of  resistance  to  compression  and  extension  be 
equal,  as  is  the  case  in  some  kinds  of  wood,  then  one  half  of 
the  fibres  will  be  compressed  while  the  other  half  are  extend- 
ed ;  and,  should  the  beam  be  of  rectangular  section,  the  neu- 
tral line  will  occur  at  the  middle  of  the  height  of  the  beam, 
and  the  condition  of  equilibrium  will  be  as  shown  in  Fig.  4- 

If  the  capability  to  resist  compression  exceeds  the  resist- 
ance to  extension,  as  in  cast-iron,  then  the  greater  portion  of 
the  fibres  will  be  employed  in  resisting  tension,  and  the  neu- 
tral line  will  be  nearer  to  the  concave  side ;  an  equilibrium 
represented  by  Fig.  6,  in  which  the  shorter  arm  of  the  lever 
may  represent  the  portion  of  the  fibres  subjected  to  compres- 
sion, and  the  longer  arm  those  suffering  tension,  and  where 
R,  the  heavier  weight,  may  represent  the  power  of  any  given 
number  of  fibres  to  resist  compression,  while  P,  the  lesser 


46  DESTRUCTIVE   ENERGY   AND    RESISTANCE.     CHAP.    III. 

weight,  represents  the  power  of  an  equal  number  of  fibres 
to  resist  tension. 

In  Art.  31  the  power  of  the  fibres  at  the  convex  side  of  a 
beam  to  resist  extension  was  shown  to  be  in  proportion  to 
the  depth  of  the  beam.  This  result  was  obtained  by  taking 
the  position  of  the  neutral  line  at  the  middle  of  the  depth. 
The  like  result  will  be  obtained  even  when  the  neutral  line 
occurs  at  a  point  other  than  the  middle.  For,  whatever  be 
the  proportionate  distance  of  this  line  from  the  lower  edge, 
that  distance,  for  the  same  material,  will  always  bear  the 
same  proportion  to  the  depth  of  the  beam. 


33.— Elements  of  Resistance  to  Rupture. — Having  now 
sufficient  data  for  the  purpose,  the  several  elements  of 
strength  which  have  been  developed  may  be  brought  to- 
gether, and  their  sum  taken  as  the  total  resistance  to  rup- 
ture. 

First. — We  have  the  rate  of  strength,  or  the  weight  in 
pounds  required  to  break  a  unit  of  the  given  material  one 
inch  square  and  one  foot  long,  when  supported  at  each  end 
(Arts.  12  and  13).  Let  ^represent  this  weight. 

Second. — We  have  the  strength  in  proportion  to  the  area 
of  cross-section,  or  to  the  product  of  the  breadth  into  the 
depth  (Arts.  16  and  17).  If  b  be  put  to  represent  the  breadth, 
and  d  the  depth,  both  in  inches,  then  this  element  of  strength 
may  be  represented  by  b  x  d  or  bd. 

Third — and  last,  we  have  the  strength  due  to  the  resist- 
ance of  the  fibres  to  a  change  in  length,  which  has  been 
shown  to  be  in  proportion  to  the  depth  (Arts.  22,  23  and 
31),  and  may  therefore  be  represented  by  d. 

Putting  these  three  elements  of  strength  together,  and 
representing  by  R  the  total  resistance,  we  have, 


DESTRUCTIVE   ENERGIES.  47 

R  =  Bxbdxd  or 

R  =  Bbd3  (5.)* 

and  this  is  the  total  power  of  resistance  to  a  cross  strain. 

34.a — Destructive  Energies. — It  is  requisite  now  to  con- 
sider the  destructive  energies.  It  has  been  shown  (Art.  27) 
that  the  power  of  a  weight,  acting  at  the  end  of  a  lever,  is 
in  proportion  to  the  length  of  the  lever.  This  is  seen  in  Fig. 
6,  where  a  small  weight  acting  at  the  end  of  the  longer  arm 
produces  as  great  an  effect  as  the  larger  weight  upon  the 
shorter  arm.  This  principle  may  be  stated  thus  :  The  mo- 
ment of  a  weight  is  equal  to  the  product  of  the  weight  into 
the  length  of  the  arm  of  leverage  at  which  it  acts. 

If  n  (Fig.  6)  be  the  arm  of  leverage,  and  P  the  weight  act- 
ing at  its  end,  then  the  moment  of  P  is  equal  to  the  weight 
P  multiolied  by  the  length  of  the  lever  n  ;  or, 

Moment  =  Pn. 

Let  5  represent  the  weight  which  it  is  found  on  trial  is 
required  to  break  a  lever  or  rod  of  given  material,  one  inch 
square,  and  projecting  one  foot  from  a  wall  into  which  it  is 
firmly  imbedded  ;  the  weight  being  suspended  from  the  free 
end  of  the  lever.  Then,  since  the  moment  equals  the  weight 
into  its  arm  of  leverage,  as  above  stated,  which  arm  in  this 
case  equals  unity,  we  have 

5  x  i  =  Pn 


*  Strictly  speaking,  the  whole  power  of  abeam  to  resist  rupture  is  due  to  the 
resistance  of  the  fibres  to  compression  and  extension, — as  will  be  shown  in  speak- 
ing of  the  resistance  to  bending — and  it  is  usual  to  obtain  the  amount  of  this 
power  by  a  more  direct  method  ;  arriving  at  the  total  resistance  by  one  opera- 
tion, and  this  based  upon  a  consideration  of  the  resistance  offered  by  each  fibre 
to  a  change  of  length,  and  taking  the  sum  of  these  resistances  ;  but  it  is  thought 
that  the  method  here  pursued  is  better  adapted  to  securing  the  object  had  in 
view  in  writing  this  work. 


48  DESTRUCTIVE   ENERGY   AND    RESISTANCE.      GHAP.  III. 

or  the  power  of  resistance  of  such  a  rod  equals  S,  the  weight 
required  to  break  it. 

Having  this  index  of  strength,  S,  and  knowing  (Art.  33) 
that  the  resistance  to  breaking  is  in  proportion  to  the  breadth 
and  the  square  of  the  depth,  then  for  levers  larger  than  one 
inch  square,  and  longer  than  one  foot,  when  the  destructive 
energy  equals  the  resistance,  we  have 

Pn  =  Sbd*  (0.) 

that  is,  for  the  moment,  or  destructive  energy,  we  have 
P,  the  weight  in  pounds,  multiplied  by  ;/,  the  length  in  feet  ; 
and  for  the  resistance,  we  have  5,  the  index  of  strength  for 
the  sectional  area  of  one  inch  square,  multiplied  by  the 
breadth  of  the  lever,  and  by  the  square  of  its  depth  ;  the 
breadth  and  depth  both  being  in  inches. 

35.  —  Rule  for  Tran§ver§e  Strength  of  Beams.  —  This  for- 
mula, (6.),  gives  a  rule  for  the  transverse  strength  of  lever?. 
From  it  we  may  derive  a  rule  for  the  transverse  strength 
of  beams  supported  at  both  ends. 

We  know,  for  example,  from  Arts.  25  and  26,  that  the 
strains  in  a  lever  are  the  same  as  in  a  beam  which  is  twice 
the  length  of,  and  loaded  at  the  middle  with  twice  the  weight 
supported  at  the  end  of  the  lever.  Therefore,  when  P  is 
equal  to  the  half  of  W,  the  weight  at  the  middle  of  a  beam 
(Fig.  5),  and  n  is  equal  to  the  half  of  /,  the  length  of  the  beam, 

we  have 

W    I      WL 
Pn  —  —  x  -  =  -  and  since,  (form.  0), 

Pn  =  Sbd*  by  substitution  we  have 

—  =Sbd*  (7.)  or 


Wl=  4Sbda  (8.) 

in  which  Wl  equals  the  moment  or  destructive  energy  of  a 
weight  at  the  middle  of  a  beam,  and  ^Sbd*  equals  the  resist- 


RULE  FOR  STRENGTH  OF  BEAMS.  49 

ance  of  the  beam.  But  this  resistance  was  found  (Art.  33) 
to  be  equal  to  Bbd2 ;  therefore, 

4$bd*  =  Bbd* 
hence  Wl  =  Bbd2  (9.) 

This  is  the  required  rule  for  the  strength  of  beams  sup- 
ported at  each  end.  In  it  W  equals  the  pounds  laid  on  at  the 
middle  of  the  beam,  /  the  length  of  the  beam  in  feet,  b  and  d 
the  breadth  and  depth  respectively  of  the  beam  in  inches, 
and  B  the  weight  in  pounds  at  the  middle  required  to  break 
a  unit  of  material  (Art.  12)  of  like  kind  with  that  in  the  beam, 
when  strained  in  a  similar  manner. 

It  may  be  observed  here  that  from 

Bbd2  —  ^Sbd2  as  above,  we  have 

B  =  4S 

or,  the  weight  at  the  middle  required  to  break  a  unit  of 
material,  when  supported  at  each  end,  is  equal  to  four  times 
the  weight  required  to  break  it  when  fixed  at  one  end  only, 
and  the  weight  suspended  from  the  other.* 


*  Professor  Moseley,  in  his  "Engineering  and  Architecture,"  puts  S to  rep- 
resent the  index  of  strength,  but  his  definition  of  this  index  shows  it  to  be  not 
the  same  as  that  for  which  S  is  put  in  this  work.  While,  with  us,  S  represents 
the  resistance  to  rupture  of  a  unit  of  material  (one  inch  square  and  one  foot 
long),  fixed  at  one  end  and  loaded  at  the  other ;  in  his  work  (Art.  408,  p.  521, 
Mahan's  Moseley,  New  York,  1856),  S  is  placed  to  represent  the  "resistance  in 
pounds  opposed  to  the  rupture  of  each  square  inch  at  the  surface  exposed  to  a  tensile 
strain" 

To  compare  the  two,  let  M  be  put  for  the  6*  of  Prof.  Moseley.  Then  his  ex- 
pression (Art.  414,  p.  528)  for  rectangular  beams, 

I      be3 

P  =  -  S  —  becomes 

6       a 

P  —  M  -r-  in  which 

da 

P  is  the  weight  at  one  end  of  a  beam,  which  is  fixed  at  the  other  end,  and  c  is 
the  depth  and  a  the  length,  both  in  inches.  If  for  c  we  put  */and  for  a  we  put  », 
representing  feet  instead  of  inches,  so  that  a  =  12  n,  then 


50  DESTRUCTIVE   ENERGY   AND    RESISTANCE.      CHAP.  III. 

36. — Formulas  Derived  from  t!ii§  Rule. — From  the  gene- 
ral formula,  (9.),  of  Art.  35,  any  one  of  the  five  quantities 
named  may  be  found,  the  other  four  being  given. 


bda 
P  -  M  -  and 


72  Pn  =  Mbd* 
Now  we  have  found  (form.  6\  that 

Pn  -  Sbd* 
Multiplying  this  by  72  gives 


72  Pn  =  TZ 

Comparing  this  value  of  72  Pn  with  that  from   Prof.  Moseley,  as  above,  we 
have 


from  which  M=-]2S 

or  M,  the  S  of  Prof.  Moseley,  is  equal  to  72  times  the  S  of  this  work. 

We  also  find  that  Prof.  Rankine  (Applied  Mechanics,  Arts.  294  and  296) 
similarly  designates  the  index  of  strength  ;  or,  as  he  and  Prof.  M.  both  term  it, 
"the  modulus  of  rupture."  Prof.  R.  defines  it  the  same  as  Prof.  M.  ;  except, 
that  instead  of  limiting  it  to  the  tensile  strain,  he  applies  it  equally  to  that  ele- 
ment, tension  or  compression,  which  first  overcomes  the  strength  of  the  beam. 

Prof.  Rankine  further  defines  it  (p.  634)  to  be  "  eighteen  times  the  load  ivhicli 
is  required  to  break  a  bar  of  one  inch  square,  supported  at  two  points  one  foot  apart, 
and  loaded  in  the  middle  between  the  points  of  support"  Now  the  bar  here  de- 
scribed is  identical  with  the  unit  of  material  adopted  in  this  work  (Arts.  12 
and  13)  ;  to  designate  the  strength  of  which  we  have  used  the  symbol  B.  To 
compare  the  two,  we  have,  as  above  found, 

M=  725 
and  also,  (Art.  35) 

B  -  4$ 

Multiplying  the  latter  equation  by  18,  we  have 

18.5  =  726"  or 

i§B  =  M  or 

as  defined  by  Prof.  Rankine,  M,  the  S  of  Prof.  Moseley,  is  equal  to  18  times  the 
value  of  B,  the  index  of  strength  as  used  in  this  work.  Hence  the  values  of 
S,  as  given  for  various  materials  by  Profs.  Moseley  and  Rankine,  are  18  times 
the  values  of  B  in  this  work  for  the  same  materials.  Owing,  however,  to  a  con- 
siderable variation  in  materials  of  the  same  name,  this  relation  will  be  found 
only  approximate. 


RULES  FOR  BREAKING  WEIGHT.  51 

For  examplej 


Bkd' 

—j-  (13.) 

Bbd* 


In  these  formulas  B  is  the  breaking  weight  in  pounds  ap- 
plied at  the  middle.  The  value  of  B  (Arts.  33  and  35)  is 
given  for  the  length  in  feet,  and  the  breadth  and  depth  in 
inches. 


QUESTIONS    FOR   PRACTICE. 


37. — What  kind  of  strain  is  a  floor  beam  subjected  to? 

38. — In  a  beam  subjected  to  a  transverse  strain,  how- 
does  the  breadth  contribute  to  its  strength  ? 

39. — How  does  the  depth  contribute  to  its  strength? 

4-0. — What  are  the  elements  of  resistance,  and  what  is  the 
expression  for  this  resistance  ? 

4-1.— When  a  beam  supported  at  each  end  carries  a  load 
at  its  middle,  what  is  the  amount  of  pressure  sustained  by 
the  two  points  of  support,  taken  together? 


52  DESTRUCTIVE   ENERGY   AND   RESISTANCE.      CHAP.  III. 

42.  —  What  portion  of  the  load  is  upheld  .by  each  sup- 
port? 

43B  —  If  the  load  be  not  at  the  middle,  what  is  the  sum  of 
the  pressures  upon  the  two  points  of  support  ? 


.  —  In  the  latter  case,  what  proportions  do  the  parts 
borne  at  the  two  points  of  support  bear  to  each  other? 

45.  —  What  expression  represents  that  borne  by  the  near 
support. 

46.  —  What  expression  represents  the  pressure  upon  the 
remote  support  ? 

47.  —  If  a  beam,  12  feet  long  between  bearings,  carries  a 
load  of  15,000  pounds,  at  a  point  4  feet  from  one  bearing, 
what  portion  of  this  load  is  borne  by  the  near  support  ? 

And  what  is  the  pressure  upon  the  remote  support? 

48.  —  When  a  beam  is  subjected  to  transverse  strain  at  its 
middle,  what  constitutes  the  destructive  energy  tending  to 
rupture? 

49.  —  When  the  destructive  energy  and  the  resistance  are 
in  equilibrium,  what  expression  represents  the  conditions  of 
the  case  ? 

50.  —  What  is  the  breaking  load  of  a  Georgia  pine  beam, 
15    feet   long  between  the   bearings;    the   breadth    being  4 
inches,  the  depth  10,  and  the  load  at  the  middle  ? 

51.  —  How  many  times  as  strong  as  when  laid  on  the  flat 
is  a  beam  when  set  on  edge  ? 


CHAPTER   IV 

THE   EFFECT   OF   WEIGHT  AS   REGARDS    ITS   POSITION. 

ART.  52. — Kelation  between  Destructive  Energy  and 
Resistance. — In  a  beam,  laid  upon  two  bearings,  and  sustain- 
ing a  load  at  the  middle,  we  have  discovered  certain  relations 
between  the  load  and  the  beam. 

The  load  has  a  tendency  to  destroy  the  beam,  while  the 
beam  has  certain  elements  of  resistance  to  this  destructive 
power. 

The  destructive  energy  exerted  by  the  load  is  equal  to 
the  product  of  half  the  load  multiplied  by  half  the  length  of 
the  beam.  The  power  of  resistance  of  the  beam  is  equal  to 
the  product  of  the  area  of  cross-section  of  the  beam,  multi- 
plied by  its  depth  and  by  the  strength  of  the  unit  of  mate- 
rial. At  the  moment  of  rupture,  the  destructive  energy  and 
the  power  of  resistance  are  equal ;  or,  as  modified  in  Art. 
35, 

Wl  =  Bbdd        or,  as  in  formula  (9.), 
WL  =  Bbd* 

53. — Dimensions  and  Weights  to  be  of  Like  Dciiomiua- 
tioiis  with  Those  of  the  Unit  Adopted. — In  applying  the  above 
formula  it  is  to  be  observed,  that  the  length',  breadth  and 
depth,  in  any  given  case,  are  to  be  taken  in  like  denominations 
with  those  of  the  unit  of  material  adopted  (Art.  33).  For  ex- 
ample :  if  the  unit  of  material  be  that  of  this  work,  then,  in 
the  application  of  the  formula,  the  breadth  and  depth  are  to 
be  taken  in  inches,  and  the  length  between  bearings  in  feet. 


54     EFFECT  OF  WEIGHT  AS   REGARDS   ITS  POSITION.    CHAP.  IV. 

It  is  also  requisite  that  the  weight  be  taken  in  like  denomi- 
nation with  that  by  which  the  resistance  of  the  unit  of  mate- 
rial was  ascertained.  If  the  one  is  in  ounces,  the  other  is 
also  to  be  in  ounces  ;  if  one  is  in  pounds,  the  other  must  be 
in  pounds ;  or,  if  in  tons,  then  in  tons. 

The  strength  of  the  unit  of  material  adopted  for  this  work 
is  given  in  pounds ;  therefore,  in  applying  the  rule,  the 
weight  given,  or  to  be  found,  must  necessarily  be  in  pounds. 

54. — Position  of  the  Weight  upon  the  Beam. — The  loca- 
tion of  the  weight  upon  the  beam  now  requires  considera- 
tion. 

Upon  our  unit  of  material,  which  is  supported  at  each 
end,  the  load  is  understood  to  have  been  located  at  the  mid- 
dle of  the  length  ;  so,  in  using  formula  (P.),  the  weight  given, 
or  sought,  must  be  located  at  the  middle  of  the  length  of  the 
given  beam. 

55. — Formula  Modified  to  Apply  to  a  L.ever.  —  By  pro- 
per modifications  this  formula  may  also  be  applied  to  the 
case  of  a  weight  suspended  from  one  end  of  a  lever  or  pro- 
jecting beam.  To  show  the  application,  we  proceed  as  fol- 
lows : 

In  Fig.  10  one  half  of  the  load  W  is  borne  on  each  one 
of  the  supports  A  and  B. 
w 


w 


FIG.  10.  FIG.  ii. 

In  Fig.  ii   we  have  a  beam  of  the  same  length,  and  sub- 
jected to  the  same  forces,  but  in  reversed  order  (Art.  26). 


BEAM   AND    LEVER   COMPARED.  55 

While  Fig.  10  represents  a  beam  supported  at  both  ends 
and  loaded  in  the  middle,  one  half  of  Fig.  n  may  be  taken 
to  represent  a  lever  projecting  from  a  wall  and  loaded  at 
the  free  end. 

In  these  two  cases  the  moment  or  destructive  energy 
tending  to  break  the  beam  is  the  same  in  each,  and  yet  it  is 
produced  in  Fig.  n  with  only  one  half  the  weight,  acting  at 
the  end  of  a  lever  only  one  half  the  length  of  the  beam.  We 
have,  therefore, 


or,  in  a  lever,  it  requires  but  a  quarter  of  the  weight  to  pro- 
duce a  given  destructive  energy,  that  is  required  in  a  beam 
of  equal  length,  laid  upon  two  supports  —  that  is  to  say,  if 
two  beams  of  like  material,  and  of  the  same  cross-section,  be 
subjected  to  transverse  strains,  in  like  positions  as  to  breadth 
and  depth,  one  beam  being  supported  at  both  ends  and  load- 
ed in  the  middle,  and  the  other  one  firmly  fixed  in  a  wall  at 
one  end  and  loaded  at  the  other  ;  and  if  the  distance  between 
the  wall  and  the  weight  in  this  latter  beam  be  equal  to  the 
distance  between  the  bearings  in  the  former  ;  then  but  one 
quarter  of  the  weight  requisite  to  break  the  beam  supported 
at  both  ends  will  be  required  to  break  the  projecting  one. 

If  the  former  requires  10,000  pounds  to  break  it,  then  the 
latter  will  be  broken  by  2500  pounds. 

The  proportion  between  the  weights  is  as  4  to  I.  But 
suppose  the  weights  upon  the  two  beams  are  equal.'  In  this 
case  the  lever  will  have  to  be  made  stronger,  and  its  sec- 
tional area  enlarged  sufficiently  to  carry  4  times  the  weight. 
Hence  we  have,  for  beams  fixed  at  one  end  and  loaded  at  the 
other, 

=  Bbd* 


in  which  W  is  the  weight  suspended  from  the   end  of  the 
lever,  and  /  is  the  length  of  the  lever  ;  or,  to  correspond  with 


56     EFFECT   OF  WEIGHT   AS   REGARDS    ITS   POSITION.    CHAP.  IV. 

the  symbols  used  in  Art.  34,  where  P  equals  the  weight  and 
;/  equals  the  length  of  the  lever,  we  have 

4/fc  =  Bbd*  (15.) 

56. — Effect  of  a  Load  at  Any  Point  in  a  Beam. — The 

next  case  for  consideration  is  that  of  the  effect  of  a  weight 
located  at  any  point  in  the  length  of  a  beam,  the  beam  being 
supported  at  both  ends. 

In  Arts.  27  and  28  it  was  shown,  in  cases  of  this  kind,  that 
R,  the  portion  of  the  whole  weight  borne  at  the  nearer  end, 

is  (form.  3.)  equal  to     Wj-\  and  that  P,  the  portion  resting 

upon   the   more    remote   end,  is    (form.  4-)    equal   to    W—,-  ; 

where  W7  equals  the  weight  on  the  beam,  R  the  portion  of  the 
weight  carried  to  the  near  support,  P  the  portion  carried  to 
the  remote  support,  /  the  length  of  the  beam,  m  the  distance 
from  the  weight  to  the  near  support,  and  n  the  distance  to 
the  remote  support. 

As  shown  in  Art.  34,  the  effective  power  or  moment  of  a 
weight  is  equal  to  the  product  of  the  weight  into  the  arm  of 
the  lever,  at  the  end  of  which  it  acts.  In  Fig.  6  the  weight 
R  may  be  taken  to  represent  the  reaction  of  the  point  of 
support  R  in  Fig.  7;  and  the  destructive  effect  at  the  point 
of  the  fulcrum  W  in  Fig.  6,  taken  to  be  the  same  as  that  at 
the  location  of  the  weight  W  in  Fig.  7,  as  the  strains  in  the 
two  pieces  are  equal ;  and  hence,  the  moment  of  R,  Fig.  6,  is 
equal  to  the  product  of  R  into  its  arm  of  lever  ;;/,  or  equal 
to  Rm. 

Taking  the  value  of  R  in  formula  (3.),  and  multiplying  it 
by  its  arm  of  lever,  /#,  we  have 


LOAD  AT  ANY  POINT  —  RULE  TESTED.          57 

Again,  taking  the  value  of  P  in  formula  (^.),  and  multi- 
plying it  by  its  arm  of  lever,  ;/,  we  have 

pn=w™-n  =  wmf 

The  two  results  agree,  as  they  should. 

57.  —  Rule  for  a  Beam  Loaded  at  Any  Point.  —  These 
formulas  may  be  tested  by  taking  the  two  extreme  condi- 
tions, the  load  at  the  middle  and  at  the  end. 

First  :    When  the  load  is  at  the  middle 

m  —  n  —  \l 
the  destructive  energy,  as  above,  will  be 

D  =  W1^  =  W  ^f=W^- 

the  same  value  as  obtained  in  Art.  35. 

Second  :  When  the  weight  is  moved  towards  the  nearer 
end,  m  becomes  gradually  shorter,  and  when  the  weight  in 
its  movement  reaches  the  point  of  support,  m  becomes  zero, 
and  n  equals  /.  The  destructive  energy  will  then  be 


as  it  ought  to  be,  for  the  weight  no  longer  exerts  any  cross 
strain  upon  the  beam. 

The  destructive  energy  therefore  of  a  weight,  W,  when 
laid  at  any  point  upon  a  beam,  is 


When  laid  at  the  middle,  it  is  as  above  shown, 


58    EFFECT  OF  WEIGHT  AS   REGARDS   ITS   POSITION.    CHAP.   IV. 

In  formula  (7.)  we  have 


therefore,  by  substitution, 

W?f  =  Sbd* 
Multiplying  by  4,  we  have 


and  since,  by  A  rt.  35, 
we  have 


=  Bbd*  (16.) 


a  rule  for  the  resistance  of  a  beam  when  the  weight  is  located 
at  any  point  in  its  length. 


58. — Effect   of  an  Equally  Distributed   Load. — Let    the 

effect  of  an  equally  distributed  weight  now  be  considered. 

Formula  (16.)  gives  the  effect  of  a  weight  at  any  point  of 
a  beam — that  is,  the  effect  of  the  weight  at  the  point  where 
it  is  located ;  but  what  effect  at  the  middle  of  the  beam  is 
produced  by  a  weight  out  of  the  middle  ? 

When  a  weight  is  hung  at  the  end  of  a  projecting  lever, 
its  effective  energy,  at  any  given  point  of  the  length  of  the 
lever,  is  equal  to  the  product  of  the  weight  multiplied  into 
the  distance  of  that  point  from  the  weight  (Art.  340. 

In  Arts.  27  and  28  we  have  the  effect  of  the  weight  W 
upon  its  points  of  support.  For  the  remote  end,  in  Fig.  7,  this 

is  P=  W  -j.     This  is  the  reaction,  or  power  acting  upward 


LOAD   AT   ANY   POINT  —  EFFECT   AT   MIDDLE.  59 

at  the  point  of  support  P.     We  have,  Arts.  56  and  57,  the 
moment  or  destructive  energy  due  to  this  reaction  equal  to 


but  if,  instead  of  the  whole  distance  ;/,  we  take  only  a  part 
of  it,  or  say  to  the  middle  of  the  beam,  or  £/,  we  have,  instead 
of/X 

Px$t  =  W~  x  \l  =  $W^-=$Wm 

or,  we  have,  for  M,  the  effect  at  the  middle  due  to  a  weight 
placed  at  any  point, 


This  result  may  be  tested  as  in  Art.  57;  for  let  m  —  |-/, 
then  M  —  |-  Wm  becomes 


which  is  a  quarter  of  the  weight  at  the  middle  into  the  whole 
length,  as  shown  in  Art.  55. 

Again,  taking  the  other  extreme  ;   when  m  becomes  zero, 
then  M  =  j-  Wm  becomes 


which  is  evidently  correct,  for  when  the  weight  is  moved 
from  over  the  clear  bearing  on  to  the  point  of  support  it 
ceases  to  exert  any  cross  strain  whatever  upon  any  point  of 
the  beam. 

From  the  above,  we  conclude  that  the  effect  produced  at 
the  middle  of  a  beam,  by  a  weight  located  at  any  point  of  its 
length,  is  equal  to  the  product  of  half  the  weight  into  its 
distance  from  its  nearest  point  of  support. 

This  result  would  be  true  of  a  second  weight,  and  a  third, 
and  of  any  number  of  weights.  If  the  weights  R,  P,  Q,  etc. 
(Fig.  12),  be  located  on  a  beam,  at  distances  from  their  near- 


60     EFFECT   OF   WEIGHT   AS   REGARDS   ITS    POSITION.    CHAP.  IV. 

cst  point  of  support  equal  to  ;;/,  r,  s,  etc.,  their  joint  effect  at 
the  middle  of  the  beam  will  be 


m  +  \Pr  +  \Qs  +  etc. 


or 


59.—  Effect  at  middle  from  an  Equally  Distributed  I,oad.—  - 

We  may  now  ascertain  the  effect  produced  at  the  middle  of 
a  beam  by  an  equally  distributed  load. 

Let  a  beam,  A  B  (Fig.  12),  of  homologous  material,  and  of 
equal  sectional  area  throughout  its  length,  be  divided   into 

any  number  of  equal  parts. 
T~|s  The  weight  of  any  one  of  these 
parts  will  equal  that  of  any 
other  part,  and  therefore  we 
have  in  this  beam  a  case  of 
an  equally  distributed  load. 
Now,  suppose  the  weight  of 

each  of  these  parts  to  be  concentrated  at  its  centre  of 
gravity,  and  represented  by  a  ball,  as  R,  P,  or  Q,  suspended 
from  that  centre  of  gravity.  Let  /  equal  the  length  of  each 

of  the  parts  into  which  the  beam  is  divided,  then  m  =  -  /, 
r  =  -  /  and  s  =  -  /,  and,  since  M  —  -  Win,  we  have  for 

22  2 

the   effect   of  the   weight   R,  at   the   middle   of   the   beam, 


=  ~R-t\  for  the  effect  of  P,  M=- 

22  2 

-      - 


-f\   and  for  the 

2 


effect  of  Q,  M  =  -  Q  -  /;  etc.,  for  all  the  weights  on  one  half 

of  the  beam. 

If  these  results  be  doubled  (for  the  effects  of  the  weights 
on  the  other  half  would  equal  these),  we  shall  have  the  total 
effect  at  the  middle  of  the  beam  of  all  the  weights.  When, 


LOAD,    CONCENTRATED   AND    DIFFUSED.  6  1 

as  in  this  case,  the  beam  is  divided  into  six  parts,  we   have 
for  the  total  effect  at  the  middle, 


-  tQ 

222 

Now  if  we  put  the  symbol  U  to  represent  a  uniformly 
distributed  load,  we  have 

^  R  =  P=Q  =  ~  therefore 


In  this  case  t  equals  —  ,  therefore 

M=^U~=l-  Ul 
468 

In  which  U  equals  the  whole  weight  uniformly  distributed 
over  the  beam. 

We  have  seen  (Art.  35)  that  \Wl  is  the  destructive  ener- 
gy of  a  weight  concentrated  at  the  centre  of  the  beam.  We 
now  see,  as  above,  that  this  same  effect  is  produced  by  \UL 

We  therefore  have 

i-£7/  —  ^Wl          or,  multiplying  by  4, 

$u=w 

or,  when  the  effects  of  the  two  loads  upon  a  beam  are  equal, 
one  half  of  £/,  the  distributed  load,  will  equal  the  load  W, 
concentrated  at  the  middle. 

60.—  Example  of  Effect  of  an  Equally  Distributed  Load. 

—Let  R,  P,  Q,  etc.,  each  equal  20  pounds;  or  the  whole  load 
U  equal  6x  20  =  120  pounds.  Let  the  whole  length,  12  feet, 
be  divided  into  six  equal  parts,  and  the  equal  loads  be  sus- 
pended from  the  centre  of  each  of  these  parts.  Then  from 
the  nearer  point  of  support,  A,  the  distance  m  to  R  is  one 
foot  ;  the  distance  r  to  P  is  three  feet  ;  and  the  distance  s  to 


62     EFFECT  OF  WEIGHT  AS   REGARDS   ITS   POSITION.   CHAP.  IV. 

Q  is  five  feet  ;  and,  since  R,  Py  and  Q  are  each  equal  to  20, 
and  (Art.  58) 

M  =    Wm  therefore 


M  =  -J-  x  20 

M  —  10  (i  +3  +  5)  —  10x9  =  90 

'  The  like  effect,  90  pounds,  is  had  from  the  three  weights 
upon  the  other  half  of  the  beam.  Adding  these,  we  have  180 
pounds.  This  is  the  destructive  energy  exerted  at  the  mid- 
dle of  the  beam  by  the  six  weights,  or  by  U,  the  120  pounds 
equally  distributed  along  the  beam.  As  a  test  of  this,  let  it 
now  be  shown  what  weight  concentrated  at  the  middle  of  the- 
beam  would  produce  the  like  effect.  In  Art.  35  we  have  for 

the  destructive  energy,  D  =  j-  Wl,  from  which   W=—     and 

¥^ 

since,  as  above,  D  =  180  and  I  =  12,  we  have   W  •=.  ---  =  60 

«J 

pounds.  This  is  the  weight  concentrated  at  the  middle. 
Above,  we  had  £/,  the  equally  distributed  Aveight,  equal  to 
1  20  pounds,  or  twice  60.  Therefore  2W  =  U.  Thus,  as  before, 
it  is  seen  that  an  equally  distributed  weight  produces  an 
effect  at  the  middle  equal  to  that  produced  by  one  half  the 
weight  if  concentrated  at  the  middle. 


6L — Rc§ult  al§o  Obtained  toy  the  Lever  Principle. — This 
result  may  also  be  obtained   by  an   application  of  the  lever 

principle.     In  Fig.  13  a  double  le- 
I  ver  is  loaded  with   weights,  pro- 
ducing strains  similar  to  those  in 
a  beam  such  as  Fig.  12.     Here  the 
arm  of  lever  at  which  R  acts  is 
five  feet,  that  of  P  three  feet,  and 
Q  one  foot ;  therefore, 


LOAD   EQUALLY  DISTRIBUTED.  63 

Rx$  =  2OX$  =  100 
Px  3  =  20  X  3  =  60 
Q  X  I  =:  20  X  I  =  20 

1  80  pounds. 

This-  is  the  whole  energy,  because  the  weights  on  the  other 
side  of  the  fulcrum  do  not  add  to  the  strain  at  W\  they  only 
balance  the  weights  R,  P,  and  Q. 

The  full  effect,  therefore,  at  the  middle  of  the  beam  is  180 
pounds,  as  before  shown,  and  this  effect  is  produced  by 
3  x  20  =  60  pounds  equally  distributed. 

Now,  what  concentrated  weight  at  the  end  of  the  lever 
would  produce  an  equal  effect  ? 

Since  the  weight  P,  at  the  end  of  a  lever,  multiplied  by 
n,  the  length  of  the  lever,  is  the  moment  or  destructive 
energy  of  the  weight,  therefore 

Pn  =  1  80  the  moment  as  above,  or 


= 
n          6 

and  this  is  one  half  of  60,  the  distributed  weight  which  pro- 
duced a  like  effect. 

Hence  we  find  that  a  given  load,  if  concentrated  at  the 
middle  of  a  beam,  will  have  a  destructive  energy  there  equal 
to  that  of  twice  said  load  equally  distributed  over  the  length 
of  the  beam  ;  or,  in  other  words,  an  equally  distributed  load 
will  need  to  be  double  the  weight  of  a  concentrated  load  to 
produce  like  effects  upon  any  given  beam. 

In  formula  (9.)  W  represents  the  concentrated  weight  at 
the  middle.  If  for  W  we  substitute  its  equivalent  %U,  we 

have 

(17.) 


64       EFFECT  OF  WEIGHT  AS  REGARDS  ITS  POSITION.      CHAP.  IV. 


QUESTIONS   FOR   PRACTICE. 


62. — A  white  pine  beam,  6x9  inches,  supported  at  each 
end,  and  set  upon  edge,  is  12  feet  long.  What  weight  laid  at 
4  feet  from  one  end  would  break  it  ? 

63. — What  weight  equally  distributed  over  the  length  of 
the  above  beam  would  break  it  ? 

64. — What  weight  concentrated  at  the  middle  of  the 
length  of  the  same  beam  would  break  it  ? 

65. — What  weight  would  break  this  beam  if  suspended 
from  one  end  of  it,  the  other  end  being  fixed  in  a  wall  ? 


CHAPTER   V. 


COMPARISON   OF   CONDITIONS — SAFE   LOAD. 

ART.  66. — Relation  between  Lengths,  Weights  and  Ef- 
fects.— In  the  consideration  of  the  effect  of  weights  upon 
beams,  we  have  deduced  certain  formulas  applicable  under 
various  conditions.  These  rules  Avill  now  be  presented  in 
such  manner  as  to  show  by  comparison :  first,  what  relation 
the  lengths  and  weights  bear  to  each  other  when  the  effects 
are  equal ;  and,  second,  the  resulting  effects  when  the  lengths 
and  weights  are  equal. 

67. — Equal  Effects. — Take  the  four  Figs.,  14,  15,  16  and  17. 

w 


FIG.  14. 


FIG.  16. 
RRRRRRRR 


FIG.  15. 


FIG.  17. 


The  lengths  of  the  beams  and  the  amounts  of  the  weights 
with  which  they  are  loaded,  are  such  as  to  produce  equal 


66  COMPARISON   OF  CONDITIONS  —  SAFE   LOAD.      CHAP.  V. 

effects.       For    example,   the   dimensions  are   such   that    in 

all  of  the  figures,  I  —  2n  and  s  —  ^  —  —?  ;  and  the  weights 

8       10 

are  so  proportioned  that   W  =  2  P  —  4  R.     By  comparison, 
we  find  that  in  Fig.  14  the  destructive  energy  is 


In  Fig.  15  the  destructive  energy  is  equal  to  the  sum  of 
the  products  of  the  several  weights  R,  into  their  respective 
distances  from  the  point  of  support  ;  or, 


Jfr(i  +  3+  5  +  7)  =  i6Rs  =  i6 
In  Fig.  16  the  destructive  energy  is 


In  Fig.  17  the  destructive  energy  equals  the  sum  of  the 
products  of  the  several  weights  R,  into  one  half  their  respec- 
tive distances  from  the  nearest  point  of  support  (Art.  58), 


or,  2 

2  [i^  (1  +  3  +  5  4-7)]  = 
16)=  i6Rs=  i 


When  the  load  is  at  any  point  upon  the  beam,  the  destruc- 

.    -J7  mil 
tive  energy  is  W  —=--. 

This  case  is  a  modification  of  Fig.  16,  for,  when 

m  —  n  =  %l  we  have, 


68.—  Comparison  of  I^en^tlis  and  Weiglit§  Producing 
Equal  Effects.  —  We  now  see  that,  in  order  to  produce  equal 
effects,  we  must  have  the  length  and  weight  in  Fig.  16  twice 


EQUAL   WEIGHTS   AND    LENGTHS.  67 

those  in  Fig.  14  ;  and  the  length  and  weights  of  Fig.  17  twice 
those  of  Fig.  15. 

Again,  we  see  that,  while  the  lengths  of  Figs.  14  and  15 
are  the  same,  the  weights  of  the  latter  are  equal  in  amount 
to  twice  that  of  the  former ;  and  that  the  same  proportions 
exist  in  Figs.  16  and  17. 


69. — Tlie  Effects  from  Equal  Weights  and  Lengths. — In 

regard  to  the  second  relation,  as  expressed  in  Art.  66. 

We  have,  in  Figs.  18,  19,  20,  and  21,  examples  showing  the 


w 


FIG.  18. 


FIG.  20. 


FIG.  19. 


FIG.  21. 


difference  of  effect  when  the  load  upon  each  beam  is  equal 
to  the  load  upon  either  of  the  other  beams,  and  the  lengths 
of  the  beams  are  equal. 

The  destructive  energy  is 

in  Fig.  18,  D  =  Pn 

19,  D 

"       20,  D 

"       21,  £> 


68  COMPARISON   OF   CONDITIONS — SAFE   LOAD.      CHAP.  V. 

70.— Rules  for  Cases  in  wliieli  the  Weights  and  Lengths 
are  Equal. — Putting  these  equal  to  the  resistance  for  levers, 
we  have  (Art.  35)  for  the  case  shown 

in  Fig.  18,  Pn  =  Sbd2 

"       19,  $Un  -  Sbd2 

"       20,  \Wl=  Sbd2 

21,  \Ul  =  Sbds 


and,  since  (Art.  35)  ^S  =  By    S  =  J/?.     If  in  the  above  we 
substitute  this  value  for  5,  we  shall  have  the  following  rules : 

For  case  i,  ^Pn  —  Bbd2  (15.) 

"     2,  2  Un  =  Bbd9  (18.) 

"     3,  Wl  =  Bbd9  (9.) 

"    4,  $Ul=  Bbd2  (17.) 

and  in  case  5,       ±W ~  =  ^^»  (.Z0.) 

this  last  being  that  of  a  load   located  at  any  point  in  the 
length  of  a  beam  (Art.  57). 


71. — Breaking  and  Safe  Loads. — These  rules  show  the 
relation  of  the  load  to  the  resistance.  Before  showing  their 
applications,  the  proportion  which  exists  between  the  break- 
ing load  and  what  is  called  the  safe  load  will  be  considered. 

72.— The    above    Rules    Useful    Only    in    Experiments.— 

The  rules  thus  far  shown  have  all  been  based  upon  the  con- 
dition of  equilibrium  between  the  destructive  power  of  the 
load  and  the  resistance  of  the  material  ;  or,  in  other  words, 
an  equilibrium  at  the  point  of  rupture.  Hence  they  are 
chiefly  useful  in  testing  materials  to  their  breaking  point. 


MARGIN    FOR   SAFETY.  69 

73. — Value  of  <r?  the  Symbol  of  Safety.  —  To  make  the 
rules  useful  to  the  architect,  it  is  requisite  to  know  what  por- 
tion of  the  breaking  load  should  be  trusted  upon  a  beam. 
It  is  evident  that  the  permanent  load  should  not  be  so  great 
as  to  injure  the  fibres  of  the  beam. 

The  proportion  between  the  safe  and  the  breaking 
weights  differs  in  different  materials.  The  breaking  load  on 
a  unit  of  material  being  represented  by  B,  as  before,  let  T 
represent  the  safe  load,  and  a  the  proportion  between  the 

r)  r) 

two  ;  or,  T  :  B  : :  i   :  a  =-™  then    T—-.     The    values  of  a, 

1  a 

for  several  kinds  of  building  materials,  have  been  found 
and  recorded  in  Table  XX.,  an  examination  of  which  will 
show  that  a,  for  many  kinds  of  materials,  is  nearly  equal  to  3, 
a  number  which  is  in  general  use.* 


74. — Value  of  «,  the  Symbol  of  Safety. — In  the  rules  a 
may  be  taken  as  high  as  we  please  above  the  value  given  for 
a  in  the  table  ;  but  never  lower  than  the  value  there  given. 

T) 

If  a  be  taken  at  4,  then,  as  above,   T  =  —  =  \B  equals  the  safe 

power  of  the  unit  of  material,  and  we  have  Wl  =  \Bbd2 ;  or, 
^.Wl^Bbd*,  as  the  proper  rule  for  a  beam  supported  at 
each  end  and  loaded  at  the  centre.  In  order,  however,  to 


*  This  is  the  value  as  fixed  by  taking  the  average  of  the  results  of  the  tests 
of  several  specimens  of  the  same  kind  of  material,  or  material  of  the  same  name. 

Owing  to  the  large  range  in  the  results  in  any  one  material,  it  is  not  safe,  in 
a  general  use  of  this  symbol,  to  take  it  at  the  average  given  in  the  table.  It 
should  for  ordinary  use  be  taken  higher. 

When  the  kind  of  material  in  any  special  and  important  work  is  known, 
and  tests  can  be  made  of  several  fair  specimens  of  it,  and  from  the  results  com- 
putations made  of  the  values  of  a,  then  an  average  of  these  would  be  safe  to  use. 
For  the  ordinary  woods  in  general  rules,  it  is  prudent  to  take  the  value  of  a  at 
not  less  than  4. 


70  COMPARISON   OF   CONDITIONS — SAFE   LOAD.      CHAP.  V. 

make  the  rules  general,  we  shall  not  adopt  any  definite  num- 
ber, as  4,  but  use  the  symbol  a.  the  value  of  which  is  to  be 
taken  from  the  table  in  accordance  with  the  kind  of  material 
employed,  increasing  its  value  at  discretion.  (Sec  note, 
Art.  73.) 

75. — Rules  for  Safe  Loaris. — The  rules,  with  this  factor  a 
introduced,  will  then  be  as  follows : 

Rule   i,  4Pan  =  Bbd*  (19.) 

"      3,  Wai  =  Bbd2  (21.) 

"      4,  \Ual-  Bbd2 


„  mn 
5,       4,Wa-j-=Bbd'  (23.) 


76. — Applications  of  tlie  Rules. — In  this  form  the  rules 
are  ready  for  use — applying  them  as  below. 

Rule  i  is  applicable  to  all  cases  where  a  load  is  suspended 
from  the  end  of  a  lever  (Fig.  18),  said  lever  being  fixed  at 
the  other  end  in  a  horizontal  position. 

Rule  2  applies  to  cases  where  a  load  is  equally  distributed 
upon  a  lever  fixed  at  one  end  (Fig.  19). 

Rule  3  is  applicable  to  a  load  concentrated  at  the  middle 
of  a  beam  supported  at  both  ends  (Fig.  20). 

Rule  4  is  applicable  to  equally  distributed  loads  upon 
beams  supported  at  both  ends  (Fig.  21). 

Rule  5  is  applicable  to  a  load  concentrated  at  any  point 
upon  a  beam  supported  at  both  ends  (Fig.  7). 

77. — Example  of  Load  at  End  of  Lever. — To  show  the 
practical  working  of  these  rules,  take,  first,  an  example 
coming  under  rule  i,  formula  (19.), 

4Pan  =  Bbd9 


LEVER  —  EXAMPLE  —  SYMBOL   OF   SAFETY.  71 

Let  it  be  required  to  find  the  requisite  breadth  and  depth  of  a 
piece  of  Georgia  pine  timber,  fixed  at  one  end  in  a  wall,  and 
•sustaining  safely,  at  five  feet  from  the  wall,  a  weight  of  1200 
pounds  ;  the  ratio  between  the  safe  and  breaking  weights 
being  taken  as  i  to  4,  and  the  value  of  B  for  Georgia  pine 
being  850  (Art.  13). 

78.  —  Arithmetical     Exemplification    of    the   Rule.  —  The 

first  thing,  in  applying  a  rule,  is  to  distinguish  between  the 
known  and  the  unknown  factors  of  an  equation,  by  so  trans- 
posing them  that  those  which  are  known  shall  stand  upon 
one  side,  and  the  unknown  upon  the  other  side  of  the  equa- 
tion. In  rule  i,  formula  (19.),  as  above,  the  known  factors 
are  4,  P,  a,  n  and  B  ;  therefore  we  transpose,  so  that 


Substituting   the  known  quantities  for  the   symbols  of  the 
first  member,  we  have 

4  x  1200  x  4  x  g 


79.  —  Caution  in  Regard  to  «,  the  Symbol  of  Safety.  —  The 

working  of  this  problem  is  interrupted  to  remark  that 
students  are  liable  to  err  in  estimating  the  value  of  a,  making 
it  a  fraction  instead  of  a  whole  number.  Thus,  if  the  pro- 
portion between  the  safe  and  the  breaking  weights  be  as 
i  to  4,  they,  starting  with  the  idea  that  the  safe  weight  is  to 
be  one  fourth  of  the  breaking  weight,  make  a  equal  to 
J,  instead  of  4.  This  is  a  serious  error,  as  the  result  would 
be  a  destructive  energy  of  only  one  sixteenth  (for  |-  :  4  :  :  i  :  16) 
of  the  true  amount,  and  consequently  the  resultant  resistance 
of  the  timber  would  be  but  one  sixteenth  of  what  it  should 


72  COMPARISON   OF   CONDITIONS—  SAFE   LOAD.      CHAP.  V. 

be,  and  in  practice  it  would  be  found  that  the  beam  would 
break  down  with  only  one  fourth  of  the  amount  considered 
the  safe  weight. 

To  farther  explain  the  value  of  a,  let  W  equal  the  break- 
ing weight,  and  T  the  safe  weight  ;  the  proportion  being  as 
4  to  i.  Then  T—  \W,  or  ^T  —  W.  Now,  in  formula  (9.) 
(Wl  —  Bbd3},  in  order  to  preserve  equality,  it  is  requisite,  in 
removing  the  symbol  W  denoting  the  breaking  weight,  that 
we  substitute  its  equal,  or  ^T.  So  when,  in  the  new  for- 
mula for  safe  weight,  W  is  understood  to  represent  not  the 
breaking  but  the  safe  weight,  ^T  becomes  ^W,  and  we  have 
^Wl  =  Bbd2  ;  therefore  the  symbol  a  is  to  be  not  a  fraction 
but  a  whole  number. 

Returning  from  this  digression  to  the  expression  at  the 
end  of  Art.  78,  and  reducing  it,  we  have 

96000 

-  =  1  12-94 


Here  we  have  the  value  of  the  breadth  multiplied  by  the 
square  of  the  depth,  but  neither  the  one  nor  the  other  is  as 
yet  determined. 

80.  —  Various  methods  of  Solving  a  Problem.  —  There 
are  at  least  three  ways  of  procedure  by  which  to  determine 
the  value  of  each  of  these  factors.  The  breadth  and  depth 
may  be  required  to  be  equal  ;  the  breadth  may  be  required 
to  bear  a  certain  proportion  to  the  depth  ;  or,  one  of  the 
factors  may  be  fixed  arbitrarily. 

First.  If  the  timber  is  to  be  square,  then  b  will  equal  d, 

bd2  =  d\  and  d   =  ty  112-94  =  4-83 

that  is,  the  dimensions  required  are  4-83,  or,  say  5  inches 
square. 


MANNER  OF  WORKING  A   PROBLEM.  73 

Second.  Let  the  breadth  be  to  the  depth  in  the  proportion 
of  6  to  10,  then 

b  :  d\  :  6  :  10  or 

iob  =  6d  or 

b  =  o-6d  Then 
112-94  =  bd2  =  o-6dx  d*  =  o-6ds 


=  d3  =  188-23  =w  that  is 


^=5-73,  and  b=  5.73x0-6  =  3-44 

The  timber  should  be  therefore  3-44  inches  broad,  and  5.73 
inches  deep  ;  or,  3^  x  5}  inches. 

Third.  The  breadth  or  depth  may  be  determined  arbitra- 
rily, or  be  controlled  by  circumstances.  Let  the  breadth  be 
fixed,  say  at  3  inches,  then 

112-94  =  bd2  —  3<af* 

;  -ii±2i=,/.  =  37-65 

d—  6-14 

The  dimensions  should  be  3  x  6-14,  or,  say,  3  x  6J  inches. 
Again,  let  the  depth  be  fixed,  say  at  6  inches,  then 

112-94  =  bd*  =  bx  6* 
112- 


thus  giving  as  the  dimensions  of  the  beam  3-  14  x  6,  or,  say 
3  J  x  6  inches. 

We   have   now   these  four  answers   to   the   question   of 
Art.  77,  namely  : 

If  the  beam  be  square,  the  side  of  the  square  must  be 
5  inches. 

If  the  breadth  and  depth  be  in  the  proportion  of  6  to  10, 
the  breadth  must  be  3!  and  the  depth  5f  inches. 


74  COMPARISON   OF   CONDITIONS — SAFE   LOAD,       CHAP.  V. 

If  the  breadth  be  fixed  at  3  inches,  then  the  depth  must 
be  6J  inches. 

If  the  depth  be  fixed  at  6  inches,  then  the  breadth  must 
be  3^  inches. 

81.— Example  of  Uniformly  Distributed   Load   on   Lever. 

— Take  an  example  coming  under  rule  2,  formula  (20.), 


Let  the  conditions  be  similar  to  those  given  in  Art.  77,  ex- 
cept that  the  weight  is  to  be  equally  distributed,  instead  of 
being  concentrated  at  the  end.     What  are  the   required   di- 
mensions of  breadth  and  depth  ? 
The  formula  transposed  becomes, 

2Uan  _ 
B 

As  the  known  factors  are  all  the  same  as  in  the  last  ex- 
ample, except  the  numerical  co-efficient,  which  here  is  only 
one  half  of  its  former  value,  it  follows  that  bd2  in  this  case 
must  be  equal  to  one  half  of  bd*  in  the  previous  case ;  or, 

- —  ='  56.47  =  bd* 


Now  to  apply  this  result  : 
First.  If  the  timber  be  square, 

56-47  =  d3  =  ^S43 

Second.  If  the  breadth  and  depth  are  to  be  as  6  to  10, 

56-47  =  0-6  d3 


and  £  =  4.55x0.6  = 


VARIOUS   LOADS   COMPARED.  75 

Third,  If  the  breadth  be  fixed  at  2  inches,  then 

56.47  =  bd3  =  2da 
^=j*  =  28.24 

2 

d=  5-31 

Fourth.  If  the  depth  be  fixed  at  5  inches,  then 

56-47  =  bd2  —  b  x  59 


The  four  answers  are,  therefore,  3-^  square  —  2f  x  4^  —  • 
2x5!  and  2  J  x  5  ;  and  the  beam  may  be  made  of  the  dimen- 
sions named  in  either  of  these  four  cases  and  be  equally 
strong. 

82.—  Load    Concentrated   at   Middle    of   Beam.  —  In     an 

example  under  rule  3,  the  value  of  bd2  in  the  formula 
Wai  =  Bbds,  would  be  just  one  quarter  of  that  required  by 
rule  i. 

83.—  Load  Uniformly  (Distributed  on  Beam  Supported  at 
Both  Ends.  —  In  cases  under  rule  4,  the  values  of  bd*  would 
be  only  one  eighth  of  those  under  rule  i  ;  and,  in  general,  the 
five  rules  given  are  so  related  that  when  the  result  of  com- 
putations under  any  one  of  them  has  been  obtained,  the  re- 
sult in  any  other  one  may  be  found  by  proportion,  in  compar- 
ing the  two  rules  applicable. 


COMPARISON   OF   CONDITIONS — SAFE   LOAD.        CHAP.   V. 


QUESTIONS  FOR   PRACTICE. 


84. — What  breadth  and  depth  are  required  for  a  white 
pine  beam,  of  sufficient  strength  to  carry  safely  3000  pounds 
equally  distributed  over  its  length,  the  beam  being  12  feet 
long  and  supported  at  each  end  ?  The  breadth  is  to  be  one 
half  of  the  depth,  and  the  factor  of  safety  a  equals  4. 

85. — What  would  be  the  size  if  square  ? 

86. — What  would  be  the  depth  if  the  breadth  be  fixed  at 
3  inches  ? 

87. — What  would  be  the  breadth  if  the  depth  were  fixed 
at  6  inches  ? 


CHAPTER  VI. 

APPLICATION   OF  RULES — FLOORS. 

ART.    88. — Application     of    Rules    to    Construction     of 

Floors.— Having  completed  the  investigation  of  the  strength 
of  beams  to  resist  rupture  so  far  as  to  obtain  formulas  or 
rules  applicable  to  the  five  principal  cases  of  strain,  we 
will  now  show  the  application1  of  these  rules  to  the  solution 
of  such  problems  as  occur  in  the  construction  of  floors.  As 
these  rules,  however,  are  founded  simply  upon  the  resistance 
to  rupture,  the  size  of  a  beam  determined  by  them  will  be 
found  to  be  much  less  than  by  rules  hereafter  given ;  and  the 
beam,  although  perfectly  safe,  will  yet  be  found  so  small  as 
to  be  decidedly  objectionable  on  account  of  its  excessive 
deflection.  Owing  to  this,  floor  beams  in  all  cases  should  be 
computed  by  the  rules  founded  upon  the  resistance  to  flexure, 
as  in  Chapter  XVII. 

89. — Proper  Rule  for  Floors. — Floor  beams  are  usually 
subjected  to  equally  distributed  loads.  For  this,  formula 
(22.)  is  appropriate,  as  it  "  is  applicable  to  equally  distributed 
loads  upon  beams  supported  at  both  ends."  It  is 

=  Bbd2 

90.— The  Load  on  Ordinary  Floors,  Equally  Distributed. — 

The  load  upon  ordinary  floors  may  be  considered  as  being 
equally  distributed  ;  at  least  when  put  to  the  severest  test— 
a  densely  crowded  assemblage  of  people.  For  this  load  all 
floors  should  be  prepared. 


7§  APPLICATION    OF   RULES  —  FLOORS.  CHAP.  VI.- 

91.  —  Floors  of  Warehouses,  Factories    and    Mills.  —  The 

floors  of  stores  and  warehouses,  factories  and  mills,  are  re- 
quired to  sustain  even  greater  loads  than  this,  but  in  all  the 
load  may  be  treated  as  one  equally  distributed. 

92B  —  Rule  for  Load  upon  a  Floor  Beam.  —  Each  beam  in 
a  floor  is  subjected  to  the  strain  arising  from  the  load  upon 
so  much  of  the  floor  as  extends  on  each  side  half  way  to  the 
next  adjoining  beam  ;  or,  that  portion  of  the  floor  which  is 
measured  by  the  length  of  the  beam  and  by  the  distance 
apart  from  centres  at  which  the  beams  are  laid.  Denote 
the  distance  apart,  in  feet,  at  which  the  beams  are  placed 
(measuring  from  the  centres'  of  the  beams)  by  c.  Then  cl 
will  equal  the  surface  of  the  floor  carried  by  one  of  the 
beams. 

If  the  load  in  pounds  upon  each  superficial  foot  of  the 
floor  be  expressed  by  /,  then  the  total  load  upon  a  floor 
beam  will  be  cfl.  This  is  an  equivalent  for  U,  the  load. 

By  substituting  for  U  its  value  cfl  in  the  formula 


we  have 

%acfl>  =  Bbd*  (24.} 

which  is  a  rule  for  the  load  upon  a  floor  beam. 

93.—  Nature  of  the  Load  upon  a  Floor  Beam.—  Before 
this  formula  can  be  used,  the  value  of  /must  be  determined. 

This  symbol  represents  a  compound  weight,  comprising 
the  weight  of  the  materials  of  construction  and  that  of  the 
superimposed  load. 

The  weight  of  the  materials  of  construction  is  also  in 
itself  a  compound  load.  A  part  of  this  load  —  the  floor  plank 
and  ceiling  (the  latter  being  either.of  boards  or  plastering)— 
will  be  a  constant  quantity  in  all  floors  ;  but  the  floor  beam 


WEIGHTS   OF    MATERIALS   OF   CONSTRUCTION.  79 

will  vary  in  weight  as  the  area  of  its  cross-section.  In  all 
cases  of  wooden  beams,  however,  the  weight  of  the  beam  is 
so  small,  in  proportion  to  the  general  load,  that  a  sufficiently 
near  approximation  to  its  weight  may  be  assigned  in  each 
case  before  the  exact  size  of  the  beam  be  ascertained. 

94-.—  Weight  of  Wooden  Beams. — For  example,  in  floors 
for  dwellings,  the  beams  will  vary  from  3x8  to  3  x  12,  ac- 
cording to  the  length  of  the  beam.  If  the  timber  be  white 
pine  (the  weight  of  which  is  about  30  pounds  per  cubic  foot, 
or  2^  pounds  per  superficial  foot,  inch  thick),  the  3x8  beam 
will  weigh  5  pounds,  and  the  3x  12  beam  7-^  pounds;  or,  as 
an  average,  say  6^  pounds  per  lineal  foot  for  all  white  pine 
beams  for  dwellings.  For  spruce,  the  average  weight  is 
about  the  same.  Hemlock,  which  is  a  little  heavier,  may  be 
taken  at  7  pounds ;  and  Georgia  pine  (seldom  used  in  dwell- 
ings) should  be  put  at  about  9  pounds  per  lineal  foot. 

95.— Weight  in  Stores,  Factories  and  Ulilfls  to  be  Esti- 
mated.— For  stores,  factories  and  mills  the  weight  is  greater, 
and  is  to  be  estimated. 

96.— Weight  of  Floor  Plank.— The  weight  of  the  floor 
plank,  if  of  white  pine  or  spruce,  is  about  3  pounds;  or,  if  of 
Georgia  pine,  about  4^  pounds  per  superficial  foot. 

97.— Weight  of  PIa§terSng. — The  weight  of  plastering 
varies  from  7  to  1 1  pounds,  and  is,  on  the  average,  about  9 
pounds,  including  the  lathing  and  furring,  per  superficial  foot. 

98.— Weight  of  Beams  in  Dwellings. — The  weights  of 
beams,  given  in  Art.  94,  are  for  the  lineal  foot,  but  it  is  re- 
quisite that  this  be  reduced  so  as  to  show  the  weight  per 
square  foot  superficial  of  the  floor.  When  the  distance  from 


8O  APPLICATION   OF   RULES — FLOORS.  CHAP.  VI. 

centres  at  which  the  floor  beams  are  placed  is  known,  the 
weight  per  lineal  foot  divided  by  the  distance  between  cen- 
tres in  feet  will  give  the  desired  result. 

Thus,  let  the  distance  from  centres  of  white  pine  floor 
beams  be  16  inches,  or  i-J-  feet.  Then  6f  -4-  i-J  =  4^  pounds. 

As  the  average  distance  from  centres  in  dwellings  differs 
little  from  16  inches,  the  weight  of  beams  may  be  safely 
taken  at  5  pounds  per  superficial  foot  for  white  pine  and 
spruce. 

99.— Weight  of  Floors  in  Dwellings. — In  summing  up 
we  have,  for  the  weight  of  the  floor  plank,  3  pounds ;  for  the 
plastering,  9  pounds,  and  for  the  beams,  5  pounds ;  and  the 
sum  of  these  items,  17,  or,  in  round  numbers,  say  20  pounds 
is  the  total  weight  of  the  materials  of  construction  upon  each 
superficial  foot  of  the  floor  of  ordinary  dwellings  ;  and  this  is 
large  enough  to  cover  the  weight  per  superficial  foot,  even 
when  a  heavier  kind  of  timber,  such  as  Georgia  pine,  is  used. 

100.— Superimposed  Load. — We  have  now  to  consider 
the  superimposed  weight,  or  the  load  to  be  carried  upon  the 
floor. 

101.— Greatest  Load  upon  a  Floor. * — Mr.  Tredgold,  in 
speaking  of  bridges,  says  (Treatise  on  Carpentry,  Art.  273): 
"  The  greatest  load  that  is  likely  to  rest  upon  a  bridge  at  one 
time  would  be  that  produced  by  its  being  covered  with  peo- 
ple." Again  he  says  :  "  It  is  easily  proved  that  it  is  about 
the  greatest  load  a  bridge  can  possibly  have  to  sustain,  as 
well  as  that  which  creates  the  most  appalling  horror  in  the 
case  of  failure."  The  floors  of  churches,  theatres,  and  other 

*  The  substance  of  the  following  discussion  of  the  load  per  foot  upon  a 
floor  was  read  by  the  author  before  the  American  Institute  of  Architects,  and 
published  in  the  Architects'  and  Mechanics'  Journal,  New  York,  in  April,  1860. 


TREDGOLD'S  ESTIMATE  OF  LOAD  ON  FLOOR.  81 

assembly  rooms,  and  also  those  of  dwellings,  are  all  liable  to 
be  covered  with  people  at  some  time  (although  not  usually), 
to  the  same  compactness  as  a  bridge.  Therefore,  to  find  the 
greatest  strain  to  which  floor  timbers  of  assembly  rooms  and 
dwellings  are  subjected,  it  will  be  requisite,  simply,  to  weigh 
the  people  ;  or,  to  find  an  answer  to  the  question  in  the  ex- 
periments of  those  who  have  weighed  them. 

102.— Tredgold's  Estimate  of  Weight  on  a  Floor. — Mr. 

Tredgold,  in  the  article  quoted,  says  :  "Such  a  load  is  about 
120  Ibs.  per  foot ;"  and  again,  at  page  283  of  his  Treatise  on  the 
Strength  of  Iron,  he  says:  "  The  weight  of  a  superficial  foot 
of  a  floor  is  about  40  Ibs.  when  there  is  a  ceiling,  counter- 
floor,  and  iron  girders.  When  a  floor  is  covered  with  peo- 
ple, the  load  upon  a  superficial  foot  may  be  calculated  at  120 
Ibs.  Therefore  120  +  40  =  160  Ibs.  on  a  superficial  foot  is  the 
least  stress  that  ought  to  be  taken  in  estimating  the  strength 
for  the  parts  of  a  floor  of  a  room." 

103.—  Tredgold's  Estimate  not  Substantiated  by  Proof.— 

Mr.  Tredgold's  most  excellent  works  on  construction  have 
deservedly  become  popular  among  civil  engineers  and  archi- 
tects. With  very  few  exceptions,  the  whole  of  the  valuable 
information  advanced  by  him  has  stood  the  test  of  the  ex- 
perience of  the  last  fifty  years  ;  and  notwithstanding  that 
many  other  works,  valuable  to  these  professions,  have  since 
appeared,  his  works  still  remain  as  standards.  Statements 
made  by  him,  therefore,  should  not  be  dissented  from  except 
upon  the  clearest  proof  of  their  inaccuracy  ;  and  only  after 
obtaining  ample  proof  is  the  statement  here  ventured  that 
Mr.  Tredgold  was  in  error  when  he  fixed  upon  120  pounds 
per  foot  as  the  weight  of  a  crowd  of  people. 

In  the  writings  of  Mr.  Tredgold,  his  positions  are  gener- 
ally sustained   by  extensive  quotations  and   references  ;  but 


82  APPLICATION   OF   RULES— FLOORS.  CHAP.  VI. 

in  this  case,  so  important,  he  gives  neither  reference,  data 
from  which  he  derives  the  result,  nor  proof  of  the  correct- 
ness of  his  statement.  This  proof  must  be  sought  else- 
where. 

104.—  Weight  of  People— Sundry  Authorities — In  the  year 
1848,  an  article  appeared  in  the  Civil  Engineer  and  Architects 
Journal,  containing  information  upon  this  subject.  From 
this  article  we  learn  that  upon  the  fall  of  the  bridge  at  Yar- 
mouth, in  May  1845,  Mr.  James  Walker,  who  was  employed 
by  government  to  investigate  the  matter,  stated  in  evidence 
before  the  coroner,  that  his  estimate  of  the  load  upon  the 
bridge  was  based  upon  taking  the  weight  of  people  at  an 
average  of  7  stone  (98  pounds)  each  ;  and  admitted  that  this 
was  a  large  estimate,  rather  higher,  perhaps,  than  it  ought 
to  be  ;  yet  he  did  so  because  it  was  customary  to  estimate 
them  at  this  weight ;  and  further,  that  he  calculated  that  six 
people  would  require  a  square  yard  for  standing  room.  At 
this  rate  there  would  be  two  persons  in  every  three  feet,  and 
the  weight  would  be  65  pounds  per  foot. 

Herr  Von  Mitis,  who  built  a  steel  suspension  bridge  over 
the  Danube,  at  Vienna,  estimated  15  men,  each  weighing  115 
Vienna  pounds,  to  a  square  fathom  of  Vienna.  This,  in  Eng- 
lish measurement  and  weight,  would  be  equal  to  39  men  in 
every  hundred  square  feet,  and  nearly  55  pounds  per  foot. 

Drury,  in  his  work  on  suspension  bridges,  lays  down  an 
arbitrary  standard  of  two  square  feet  per  man  of  10  stone 
weight.  This  equals  70  pounds  per  superficial  foot. 

In  testing  new  bridges  in  France,  it  is  usual  for  govern- 
ment to  require  that  200  kilogrammes  per  square  metre  of 
platform  shall  be  laid  on  the  bridge  for  24  hours.  This  is 
equal  to  41  pounds  per  foot. 

The  result  of  combining  the  above  four  instances  is  an 
average  of  57!  pounds  per  foot. 


WEIGHT   OF   PEOPLE.  83 

But  we  have  a  more  accurate  estimate,  founded  upon 
trustworthy  data.  Quetelet,  in  his  Treatise  on  Man,  gives  the 
average  weight  of  males  and  females  of  various  ages  as 
follows  : — 

Average  weight  of  males  at    5,  10  and  15  years,      61-53 

"  20      "     25        "  I35-59 

"  "  30,  40      "      50        "  140.21 

Average  weight  of  females  at    5,  10  and  15  years,      57-50 

20    "    25      "         116-33 
"  "  "  30,40    "    50      "         121-80 


6J  632-96 


Total  average  weight  in  Ibs.  —  105-5 


105.— Estimated  Weight  of  People  per  Square  Foot  of 
Floor. — The  weight  of  men,  women  and  children,  therefore, 
is  105.5  pounds  each,  on  the  average.  This  may  be  taken  as 
quite  reliable  as  to  the  weight  of  people.  Now  as  to  the 
space  occupied  by  them. 

It  is  known  among  military  men  that  a  body  of  infantry 
closely  packed  will  occupy,  on  the  average,  a  space  measur- 
ing 15  x  20  —  300  square  inches  each.  At  this  rate,  48  men 
would  occupy  100  square  feet,  and  if  a  promiscuous  assembly 
should  require  the  same  space  each,  then  there  would  be  a 
load  of  50-64  pounds  upon  each  square  foot.  In  military 
ranks,  however,  the  men  would  weigh  more.  Taking  the 
weight  of  males  from  20  to  50  years,  in  the  above  table — 
this  being  the  probable  range  of  the  ages  of  soldiers — the 
average  is  found  to  be  137-9;  a  weight  of  66  pounds  upon 
each  superficial  foot  of  floor  ;  and  this  weight  may  be  taken 
as  the  greatest  which  can  arise  from  a  crowd  of  people. 


84  APPLICATION    OF   RULES— FLOORS.  CHAP.  VI. 

106.— Weight   of  People,  Estimated  a§   a    Uve  Load.— 

But  this  is  simply  the  weight,  no  allowance  being  made  for 
any  increase  of  strain  by  reason  of  the  movement  of  the 
people  upon  the  floor.  We  will  now  consider  the  increase 
made  in  consequence  of  the  agitation  of  the  weight  through 
walking  and  other  movements. 

In  walking,  the  body  rises  and  falls,  producing  in  its  fall 
a  strain  additional  to  that  due  to  its  weight  when  quiet. 

The  moving  force  of  a  falling  body  is  known  to  be  equal 
to  the  square  root  of  64^  times  the  space  fallen  through  in 
feet,  multiplied  by  the  weight  of  the  body  in  pounds.  By 
this  rule,  knowing  the  weight  and  the  height  of  fall,  we  may 
compute  the  force. 

The  weight  in  the  present  case,  66  pounds,  is  known,  but 
the  height  of  fall  is  to  be  ascertained.  This  height  is  not 
that  of  the  rise  and  fall  of  the  foot,  but  of  the  body  ;  the  latter 
being  less  than  the  former.  The  elevation  of  body  varies 
considerably  in  different  persons,  as  may  be  seen  by  observ- 
ing the  motions  of  pedestrians.  Some  rise  and  fall  as  much 
as  half  an  inch  at  each  step,  while  others  deviate  from  a 
right  line  but  slightly.  If,  in  the  absence  of  accurate  obser- 
vation, the  rise  be  assumed  at  a  quarter  of  an  inch,  as  a  fair 
average,  then  the  moving  force  of  the  66  pounds,  computed 
by  the  above  rule,  would  be  76.4  pounds.  This  would  be  the 
moving  force  at  the  moment  of  contact,  and  the  effect  pro- 
duced would  be  equal  to  this,  provided  that  the  falling  body 
and  the  floor  were  both  quite  inelastic ;  but  owing  to  the 
presence  of  an  elastic  substance  on  the  soles  of  the  feet,  and 
at  the  joints  of  the  limbs,  acting  as  so  many  cushions,  the 
force  of  the  blow  upon  the  floor  is  much  diminished.  The 
elasticity  of  the  floor  also  diminishes  the  effect  of  the  force 
to  a  small  degree.  Hence  the  increase  of  over  ten  pounds, 
as  found  above,  would  be  much  diminished,  probably  one 
half,  or,  say  to  six  pounds. 


ACTUAL   WEIGHT   OF    MEN.  85 

I07L—  Weight  of  Military. — This  six  pounds  would  be  the 
increase  per  foot  superficial.  To  make  this  effect  general  over 
the  whole  surface  of  the  floor,  it  is  requisite  that  the  weight 
over  the  whole  surface  fall  at  the  same  instant ;  or,  that  the 
persons  covering  the  floor  should  all  step  at  once,  or  with 
regular  military  step.  It  will  be  found  that  this  is  the  se- 
verest test  to  which  a  floor  of  a  dwelling  or  place  of  assem- 
bly can  be  subjected.  In  promiscuous  stepping  the  strain 
would  be  much  less,  scarcely  more  than  the  quiet  weight  of 
the  people. 

108.— Actual  Weights  of  men  at  Jackson's  and  at  Hoes' 
Foundries. — The  above  results,  it  must  be  admitted,  are  de- 
rived from  data — with  reference  to  the  height  of  fall,  and  to 
the  lessening  effect  of  the  elastic  intervening  substances, — 
which  are  in  a  measure  assumed,  and  hence  are  not  quite 
conclusive.  They  need  the  corroboration  of  experiment. 

To  test  them,  I  experimented,  in  April,  1860,  at  the  foun- 
dry of  Mr.  James  L.  Jackson  in  this  city.  He  kindly  placed 
at  my  service  his  workmen  and  his  large  scale.  The  scale 
had  a  platform  of  8^  x  14  feet.  It  was  of  the  best  construc- 
tion, and  very  accurate  in  its  action.  Eleven  men,  taken 
indiscriminately  from  among  the  workmen  of  the  foundry, 
stood  upon  the  platform.  Their  combined  weight  while 
standing  quietly  was  1535  pounds,  being  an  average  of  139-55 
pounds  per  man.  This  is  but  a  pound  and  a  half  more  than 
was  derived  from  the  tables  of  Quetelet.  It  is  quite  satisfac- 
tory in  substantiating  the  conclusion  there  drawn.* 


*  In  May,  1876,  since  the  above  was  written,  by  the  courtesy  of  Messrs.  R. 
Hoe  &  Co.,  of  this  city,  who  placed  at  my  disposal  their  platform  scale  and  men, 
I  was  enabled,  by  a  second  experiment,  to  ascertain  the  weight  of  men  and  the 
space  they  occupy.  Selecting  twenty-six  stalwart  men  from  their  smith  shop, 
they  were  found  to  weigh  3955  pounds,  and  to  occupy  upon  the  platform  a 
space  7  x  *1\  =  S2^  square  feet,  or  753-  pounds  per  superficial  foot.  This  is  a 


86  APPLICATION   OF   RULES — FLOORS.  CHAP.  VI. 

109.— Actual   measure  of  Live  Load. — After  ascertaining 

the  quiet  weight  of  the  men,  they  commenced  walking  about 
the  platform,  stepping  without  order,  and  indiscriminately. 
The  effect  of  this  movement  upon  the  scale  was  such  as  to 
make  it  register  1545  pounds ;  an  increase  of  only  ten  pounds, 
or  less  than  one  per  cent.  They  were  then  formed  in  a  circle 
and  marched  around  the  platform,  stepping  simultaneously 
or  in  military  order.  The  effect  upon  the  scale  produced  by 
this  movement  was  equal  to  1694  pounds,  an  increase  of  159 
pounds,  or  over  ten  per  cent.  This  corroborated  the  results 
of  the  computation  before  made  most  satisfactorily  ;  ten  per 
cent  of  the  weight  per  foot,  66  pounds,  being  6-6  pounds. 

As  a  final  trial,  the  men  were  directed  to  use  their  utmost 
exertion  in  jumping,  and  were  urged  on  in  their  movements 
by  loud  shouting.  The  greatest  consequent  effect  produced 
was  2330  pounds,  an  increase  of  795  pounds,  or  about  52  per 
cent. 

110.— More  Space  Required  for  Live  Load. —  This  seems 
a  much  more  severe  strain  than  the  former,  but  we  must 
consider  that  men  engaged  in  the  violent  movements  neces- 
sary to  produce  this  increase  of  over  50  per  cent  need  more 
standing  room.  Packed  closely,  occupying  only  15  x  20 
inches  (the  space  allowed  per  man  in  computing  the  weight 
per  foot  to  be  66  pounds),  it  would  not  be  possible  to  move 
the  limbs  sufficiently  for  jumping.  To  do  this,  at  least 
twice  as  much  space  would  be  required.  But,  to  keep  within 
the  limits  of  safety,  let  only  one  half  more  space  be  allowed. 
In  this  case  the  66  pounds  would  be  the  weight  on  a  foot 


larger  average  than  found  at  Mr.  Jackson's,  or  than  any  previous  weight  on 
record,  and  is  accounted  for  by  the  fact  that  these  were  muscular  men,  weighing 
about  12^  pounds  each  more  than  the  heaviest  hereinbefore  noticed,  and  much 
heavier  than  it  were  reasonable  to  expect  in  assemblages  generally. 


MEASURE   OF   LIVE   LOAD.  S/ 

and  a  half,  or  there  would  be  but  44  pounds  on  each  foot  of 
surface. 

Add  to  this  the  50  per  cent  for  the  effects  of  jumping,  or 
22  pounds,  and  the  sum,  66  pounds,  is  the  total  effect  of  the 
most  violent  movements  on  each  foot  of  the  floor  ;  the  same 
as  for  the  weight  of  men  standing  quietly,  but  packed  so 
much  more  closely. 

III.— 3fo  Addition  to  Strain  by  Live  Load. — The  greatest 
effect,  then,  that  it  appears  possible  to  produce  by  an  assem- 
bly on  a  floor,  is  from  the  regular  marching  of  a  body  of  men, 
closely  packed  ;  and  amounts  to  66  +  6-6  =  72-6  pounds  per 
superficial  foot. 

This  result  would  show  the  necessity  of  providing  for  ten 
per  cent  additional  to  the  weight  of  the  people.  This  in 
general  is  not  needed,  for  the  conditions  of  the  case  generally 
preclude  the  possibility  of  obtaining  this  additional  strain 
upon  the  floor.  The  strain  of  66  pounds  is  only  obtained  by 
crowding  the  people  closely  together  in  the  whole  room. 
To  obtain  the  ten  per  cent  additional  strain,  they  must  be  set 
to  marching  ;  but  there  is  no  space  in  which  to  march,  unless 
they  march  out  of  the  room,  and  in  doing  this  the  strain  is 
not  increased,  for  the  weight  of  those  who  pass  out  is  fully 
equal  to  the  stress  caused  by  the  act  of  marching. 

Were  both  ends  of  the  room  quite  open,  or  were  it  a  long 
hall,  as  a  bridge,  through  which  the  people  could  march 
solid,  the  throng  being  sufficiently  numerous  to  keep  the  floor 
constantly  full,  then  the  ten  per  cent  would  need  to  be  added, 
but  not  in  ordinary  cases  of  floors  of  rooms. 

112.—  Margin  of  Safety  Ample  for  Momentary  Extra  Strain 
in  Extreme  Ca§es. — It  may  be  argued  still,  that,  although  the 
room  be  full  and  marching  can  only  be  effected  by  some  of 
the  people  leaving  the  floor,  yet  this  additional  strain  will  be 


88  APPLICATION   OF   RULES  —  FLOORS.  CHAP.  VI. 

obtained  in  consequence  of  the  exertion  made  in  the  act  of 
taking  the  very  first  step,  before  any  have  left  the  room. 
To  this  we  reply  that  the  strain  thus  produced  would  not 
endanger  the  safety  of  the  floor,  because  this  strain,  when 
compared  with  the  ultimate  strength  of  the  beams  sustaining 
it,  would  be  quite  small,  and  its  existence  be  but  momentary. 
Beams  made  so  strong  as  not  to  break  with  less  than  from 
three  to  five  times  the  permanent  load  would  certainly  not 
be  endangered  by  the  addition  for  a  moment  of  only  ten  per 
cent  of  that  load. 

113.—  Weight  Reduced  by  Furniture  Reducing  Standing 
Room.  —  Hence,  for  all  ordinary  cases,  no  increase  of  strength 
need  be  made  for  the  effects  of  motion  in  a  crowd  of  people 
upon  a  floor,  and  therefore  the  amount  before  ascertained, 
66  pounds,  or,  in  round  numbers,  say  70  pounds,  may  be  used 
in  the  computations  as  the  full  strain  to  which  the  beams 
may  be  subjected.  Indeed,  the  cases  are  rare  where  the 
strain  will  even  be  as  much  as  this.  When  we  consider  the 
space  occupied  in  dwellings  by  furniture,  and  in  assembly 
rooms  by  seats,  the  presence  of  these  articles  reducing  the 
standing  room,  the  average  weight  per  foot  superficial  will 
be  found  to  be  very  much  less. 


114.—  The  Greatest  Load  to  be  provided  for  i§  ?O  Pounds 
per  Superficial  Foot.  —  As  a  conclusion,  therefore,  floor  beams 
computed  to  safely  sustain  70  pounds  per  superficial  foot,  or 
to  break  with  not  less  than  three  or  four  times  this,  will  be 
quite  able  to  bear  the  greatest  strain  to  which  they  may  be 
subjected  in  the  floors  of  assembly  rooms  or  dwellings;  and 
especially  so  when  the  precaution  of  attaching  them  to  each 
other  by  bridging*  is  thoroughly  performed,  thereby  ena- 


*  The  subjects  of  Floor  Beams  and  of  Bridging  are  farther  treated  in  Chap- 
ters XVII.  and  XVIII. 


RULE    FOR   FLOORS   OF   DWELLINGS.  89 

bling  the  connected  series  of  beams  to  sustain  the  concen- 
trated weight  of  a  few  heavier  persons  or  of  some  heavy 
article  of  furniture. 

MS.— Rule  for  Floors  of  Dwellings. — We  now  have,  by 
including  the  weight  of  the  materials  of  construction  as 
shown  in  Art.  99,  the  total  weight  per  superficial  foot,  as 

follows  :— 

/  =  70  +  20  =  90 

for  the  floors  of  dwellings.    With  this  value  of/,  formula  (&4-\ 

i-  acfl2  =  Bbd*  becomes 

•J-  ac  90  /*  =  Bbd3  or,  when  a  =  4 

i$ocl'=Bbd*  (25) 

116. — Distinguishing  Between  Known  and  Unknown  Quan- 
tities.— This  formula  may  now  be  applied  in  determining 
problems  of  floor  construction  in  dwellings,  in  which  the  safe 
strength  is  taken  at  one  fourth  of  the  breaking  strength. 

In  distinguishing  between  known  and  unknown  quantities, 
we  will  find  generally  that  B  and  /  are  known,  while  c,  b  and  d 
are  unknown. 

From  formula  (25.)  therefore,  we  have,  by    grouping  these 

quantities, 

_L8oT=^  (M-) 

(17.— Practical  Example. — Formula  (26.)  is  a  general  rule 
for  the  strength  of  floor  beams  of  dwellings. 

As  an  example  under  this  rule,  let  it  be  required  to  find 
the  sectional  dimensions  and  the  distance  from  centres  of  the 
beams  of  a  floor  of  a  dwelling;  the  span  or  length  between 
bearings  being  20  feet,  and  the  material,  spruce. 


90  APPLICATION   OF   RULES—  FLOORS.  CHAP.  VI. 

Here  B  =  550  and  /=  20;  and  from  formula  (26.) 

i8ox2o2        bd* 

-  =  ---  =  1300 
550  c 

118.—  Eliminating  Unknown  Quantities  —  We  have  here 
the  numerical  value  of  a  quotient,  arising-  from  a  division  of 
the  product  of  the  breadth  and  square  of  the  depth,  by  the 
distance  from  centres  at  which  the  beams  are  to  be  placed. 

Two  of  the  three  unknown  quantities  are  now  to  be  as- 
signed a  value,  before  the  third  can  be  determined.  Circum- 
stances will  indicate  which  two  may  be  thus  eliminated.  In 
some  cases  the  breadth  and  depth  of  the  timber  arc  fixed, 
and  the  distance  from  centres  is  the  unknown  quantity  ; 
in  others,  the  distance  from  centres  and  the  depth  may  be  the 
fixed  quantities,  and  the  breadth  be  the  factor  to  be  found  ; 
or,  the  distance  and  the  breadth  be  fixed  upon,  and  the  depth 
be  the  quantity  sought  for.  Generally,  the  breadth  and 
depth  are  assigned  according  to  the  requirements  of  the  case, 
or  simply  as  a  trial  to  ascertain  the  scope  of  the  question,  and 
the  distance  from  centres  is  the  dimension  left  to  be  deter- 
mined by  the  formula. 

119.  —  Isolating  the  Required  Unknown  Quantity.  —  In  the 

solving  of  a  question  of  either  kind,  the  formula  must  first 
be  transposed  so  as  to  remove  all  of  the  factors,  except  the 
one  sought  for,  to  the  same  side  of  the  equation  ;  thus, 

bd*  ,  .,, 

—  =  130-9  becomes  either 

or 


c  = 


b 
bd2 


130-9 

Assuming  the  value  of  any  two  of  the  factors,  we  select  the 
proper  formula  and  proceed  with  the  test  for  the  third  factor. 


RULE   FOR   DISTANCE   FROM    CENTRES.  9! 

(20.—  distance  from  Centres  at  Given  Breadth  and  Depth. 

For  example,  fix  the  breadth  and  depth  at  3  and  9  inches. 
Then  to  find  c,  the  above  expression, 

W* 

c  =  —  becomes 

130-9 


130-9       130-9 

The  value  of  c  being  in  feet,  this  gives  about  i  foot  10  inches, 
or  22  inches. 

121.—  Distance  from  Centres  at  Another  Breadth  and 
Depth.  —  The  above  result  may  be  considered  too  great,  and 
beams  of  less  size  and  nearer  together  be  more  desirable. 
If  so,  assume  a  less  size,  say  3x8;  we  then  have 

3  x  82        102 

c  —  ^-  -=i-47 

130-9       130-9 

This  gives  c  equal  to  about  \*j\  inches. 

122.—  Distance  from  Centres  at  a  Third  Breadth  and 
Depth.  —  With  the  object  in  view  of  economy  of  material,  let 
another  trial  be  had,  fixing  the  size  at  2\  x  9.  In  this  case 

,=  2*X9!  =  202-5   =  I.S5 
130-9          130-9 

This  gives  for  c  about  i8J  inches.     The  answers  then  to  this 
problem  are, 

for  3  x  9  inches,  22  inches  from  centres, 

"3x8      "  \j\ 

and    "  2j  x  9      "  i8£         "  " 

These  trials  may  be  extended  to  any  other  proportions 
thought  desirable,  fixing  first  the  breadth  and  depth,  and 
then  determining  the  corresponding  value  of  c.  (See  pre- 
caution, Art.  88.) 


92  APPLICATION   OF   RULES  —  FLOORS.  CHAP.  VI. 

123.—  Breadth,  the  Depth  and  Distance  from  €entre§ 
being  Given.  —  Again,  it  may  be  desirable  to  assume  a  value 
for  c,  and  then  to  ascertain  the  proper  corresponding  breadth 
and  depth.  In  this  case,  one  of  the  two  unknown  factors,  b 
and  d,  must  also  be  assumed.  Let  us  fix  upon  c  =  1-5  and 
d  —  8,  then  the  formula  in  Art.  119, 


c  l3O-9  x  1-5 

becomes    b  —  -  ^  =  3.07 


j  -  VJ  \~>  V>  \J  i  -L 1  \~>  O  t/     —  ^ 

or,  say  3  inches  for  the  breadth. 

1 2 4-. — Depth,  the    Breadth    and    Distance    from    Centres 

being  Given. — If  the  breadth  be  assumed,  say  at  2^,  then, 
with   <;  =  1.5,    to  find  the  depth  we  have  (Art.  119), 


d=  8-86  =  8|  inches. 

Thus,  when  placed  at   18   inches  from   centres,  we  have,  in 
the  one  case  3x8  inches,  and  in  the  other  2^  x  8J  inches. 

125.— General  Rules  for  Strength  of  Beams. — Any  other 
case  of  wooden  beams  for  dwellings  may  be  treated  in  a 
similar  manner,  using  formula 


Beams  of  any  material  for  any  building  may  be  deter 
mined  by  the  general  formula 


in  all  cases  regarding  the  caution  given  in  Art.  88, 


QUESTIONS    FOR    PRACTICE. 


126. — In  the  floor  of  a  dwelling,  composed  ot  3  x  9  inch 
beams  16  feet  long,  how  far  from  centres  should  spruce 
beams  be  placed  ? 

127. — How  far  if  of  hemlock? 

• 

128. — How  far  if  of  white  pine? 

129. — In  the  floor  of  a  dwelling,  composed  of  2j  x  10  inch 
beams  19  feet  long,  •  how  far  from  centres  should  spruce 
beams  be  placed? 

130. — How  far  if  of  hemlock? 
1 3  I. — How  far  if  of  white  pine  ? 

132. — In  a  floor  of  4  x  12  inch  beams  23  feet  long,  and  re- 
quired to  carry  150  pounds  per  superficial  foot  (including 
material  of  construction),  how  far  from  centres  should  spruce 
beams  be  placed,  the  factor  of  safety  being  4? 

133. — How  far  if  of  hemlock? 
134. — How  far  if  of  white  pine  ? 
135. — How  far  if  of  Georgia  pine? 


CHAPTER   VII. 

GIRDERS,    HEADERS   AND    CARRIAGE   BEAMS. 

ART.  I360 — A  Girder  Defined. — By  the  term  girder  is 
meant  a  heavy  timber  set  on  posts  or  other  supports,  and 
serving,  as  a  substitute  for  a  wall,  to  carry  a  floor. 

137. — Rule  for  Girders. — A  girder  sustaining  a  tier  of 
floor  beams  carries  an  equally  distributed  load  ;  the  same  per 
superficial  foot  as  that  which  is  carried  by  the  floor  beams. 
In  determining  the  size  of  the  girder  formula  (24-)  is  appli- 
cable, namely, 

\acfl*  =  Bbd2 

138. — Distance  between  Centres  of  Girders. — In  apply- 
ing this  formula  to  girders,  it  is  to  be  observed  that  c  repre- 
sents the  distance  between  centres  of  girders,  Avhere  there 
are  two  or  more,  set  parallel ;  or,  the  distance  from  centre  of 
girder  to  one  of  the  walls  of  the  building,  if  the  girder  be 
located  midway  between  the  two  walls ;  or,  an  average  of 
the  two  distances,  if  not  midway.  As  an  example  of  the 
latter  case, — in  a  building  30  feet  wide,  the  centre  of  a  girder 
is  12  feet  from  one  wall  and  18  feet  from  the  other.  Here 


GIRDERS  —DISTANCE   BETWEEN.  95 

139.  —  Example  of  Distance  from  Centres.  —  What    is    the 
required  size  of  a  Georgia  pine  girder  placed  upon  posts 
set   15   feet  apart,  the  centre  of  the  girder  being  12  feet  from 
one  wall  and    18   feet  from  the  other;  the  load  per  foot  super- 
ficial of  floor,  including  the  weight  of  the  materials  of  con- 
struction, being  100  pounds,  and  the  value  of  a    being  taken 
at  4? 

140.  —  §ize  of  Girder  Required  in  above  Example.  —  By 

transposing  formula  (@4-)  we  have 


B 

and  if  the  breadth  be  to  the  depth  in  the  proportion  of,  say 
7  to    10,    then  (Art.  80) 

*.  =     • 


0-5  XAX  15  x  100  x  15*          .. 

—  5  --  £_  —  d*  —  1134-45 
07x850 


d  =  y  U34-45  -  10-43 

and   b  =  0-7  x  10-43  =  7'3°- 

Therefore  the  girder  should  be  7-3  x  10-43  ;  or>  to  avoid 
fractions,  say  8  x  n  inches. 

14-1. — Framing   for   Fireplaee§,  Stairs    and  Light-wells.— 

We  will  now  consider  the  subject  of  framing  around  open- 
ings in  floors,  for  fireplaces,  stairs  and  light-wells. 

142.—  Definition  of  Carriage  Beams,  Headers  and  TaiB 
Beams. — Fig.  22  may  be  taken  for  a  representation  of  a  stair- 
way opening  in  a  floor  ;  AB  and  CD  being  the  walls  of  the 


96  GIRDERS,    HEADERS   AND   CARRIAGE   BEAMS.      CHAP.  VII, 


FIG.  22. 

building,  A  C  and  BD  the  carriage  beams  or  trimmers,  EF  the 
header,  and  the  beams  which  reach  from  the  header  to  the 
wall  CD,  such  as  GH,  the  tail  beams. 

14-3.— Formula   for   Headers— General    Considerations.— 

First,  the  headers. 

The  load  upon  the  header  EF  is  equally  distributed, 
therefore  formula  (22.)  is  applicable. 

%Ual—  Bbd2 

The  header  carries  half  the  load  upon  the  tail  beams,  or 
the  load  upon  a  space  equal  to  the  length  of  the  header  by 
half  the  length  of  the  tail  beams.  Let  g  represent  the  length 
of  the  header,  n  the  length  of  the  tail  beams,  and  /  the  load 
per  foot  superficial ;  then  £/,  the  load  upon  the  header,  equals 


and,  as  g  here  represents   /,    the  length,  therefore, 


HEADERS—  RULE  —  PRECAUTION.  97 

and  formula  (22.)  becomes 


=  Bbd* 
lafng2  =  Bbd3 

I44B  —  Allowance  lor  Damage  by  Mortising.  —  This  last 
formula  should  be  modified  so  as  to  allow  for  the  damage 
done  to  the  header  by  the  mortising-  for  the  tenons  of  the 
tail  beams.  This  cutting  of  the  header  ought  to  be  confined 
as  nearly  as  possible  to  the  middle  of  its  height,  so  that  the 
injury  to  the  wood  may  be  at  the  place  where  the  material 
is  subject  to  the  least  strain. 

If  this  is  properly  attended  to;  it  will  be  a  sufficient 
modification  to  make  the  depth  of  the  header  one  inch  more 
than  that  required  by  the  formula.  Thus,  when  the  depth 
by  the  formula  is  required  to  be  9  inches,  make  the  actual 
depth  10;  or,  for  d*  substitute  (d—ij,  d  being  the  actual 
depth.  The  rule,  thus  modified,  will  determine  a  header  of 
the  requisite  strength  with  a  depth  one  inch  less  than  the 
actual  depth.  This  will  compensate  for  the  damage  caused 
by  mortising. 

The  expression  in  the  last  article  then  becomes 

\afng>  =,  Bb(d-lJ 

14-5.  —  Rule  for  Headers.  —  Generally,  the  depth  of  a 
header  is  equal  to  the  depth  of  the  floor  in  which  it  occurs. 
Hence,  when  the  depth  of  the  floor  beams  has  been  deter- 
mined, that  of  the  header  is  fixed.  There  remains  then  only 
the  breadth  to  be  found. 

We  have,  for  the  breadth  of  a  header  (from  Art.  144) 

afng* 

~- 


(See  precaution  in  Art.  88.) 


98  GIRDERS,    HEADERS   AND   CARRIAGE   BEAMS.     CHAP.  VII. 


Example.  —  In  a  tier  of  nine  inch  beams,  what  is 
the  required  breadth  of  a  white  pine  header  at  the  stair- 
way of  a  dwelling,  the  header  being-  12  feet  long,  and  carry- 
ing tail  beams  16  feet  long;  the  factor  of  safety  being  4? 

In  formula  (27.),  making  a  =  4,  /=  90,  n=  16,  g—  12, 
B  =  500  and  d  —  9,  the  formula  becomes 

4  x  90  x  1  6  x  1  2  2 

b  =  -  TOT-  =  6-48 

4  x  500  x  8 

The  breadth  of  the  header  should  be  6^,  or  say  7  inches, 
and  its  size  7x9  inches. 

14-7.  —  Carriage  Beam§  and  Bridle  Irons.  —  A  carriage 
beam,  or  trimmer,  in  addition  to  the  load  of  an  ordinary 
beam,  is  required  to  carry  half  the  load  of  the  header  which 

hangs  upon  it  for 
support.  As  this  is 
a  concentrated  load 
at  the  point  of  con- 
nection, all  mortising 
at  this  point  to  re- 
ceive the  header 
FIG.  23.  should  be  carefully 

avoided,  and  the  requisite  support  given  with  a  bridle  iron, 
as  in  Fig.  23. 

148.  —  Rule  for  Bridle  Iron*.  —  In  considering  the  strain 
upon  a  bridle  iron,  we  find  that  it  has  to  bear  half  the  load 
upon  the  header,  and,  as  the  iron  has  two  straps,  one  on  each 
side  of  the  header,  each  strap  has  to  bear  only  a  quarter  of 
the  load  upon  the  header. 

We  have  seen  (Art.  14-3)  that  the  load  upon  the  header 
equals  %fng,  where  g  represents  the  length  of  the  header,  n 
the  length  of  the  tail  beams,  both  in  feet,  and  /  the  load  per 


BRIDLE    IRONS  —  RULE.  99 

superficial  foot.      The  load   upon   each  strap  of  the  bridle 
iron  will,  therefore,  be  equal  to 


Good  refined  iron  will  carry  safely  from  9000  to  15,000 
pounds  to  the  square  inch  of  cross-section.  Owing,  however, 
to  the  contingencies  in  material  and  workmanship,  it  is  pru- 
dent to  rate  its  carrying  power,  for  use  in  bridle  irons,  at  not 
over  9000  pounds. 

If  the  rate  be  taken  at  this,  and  r  be  put  to  represent  the 
number  of  inches  in  the  cross-section  of  one  strap  of  the 
bridle  iron,  then  9000^  equals  the  pounds  weight  which  the 
strap  will  safely  bear;  and  when  there  is  an  equilibrium  be- 
tween the  weight  to  be  carried  and  the  effectual  resistance, 
we  shall  have 

\fng- 


from  which  r  = 


72000 

(4-9.—  Example.  —  For  an  example,  let  f  =  100,  n  —  16, 
and  g=  12  ;  then 

100  x  16  x  12 

r  —  -  -  —  0-266 

72000 

If  the  bridle  iron  were  made  of  J  by  i^  inch  iron 
(-J-x  i£  =  0-375)  tne  SIZG  would  be  ample.  For  such  a  header 
they  are  usually  made  heavier  than  this,  yet  this  is  all  that  is 
needed.  It  is  well  to  have  the  bridle  iron  as  broad  as 
possible,  in  order  to  give  a  broad  bearing  to  the  wood,  so 
that  it  shall  not  be  crushed. 

ISO.  —  Rule  for  Carriage  Beam  with  One  Header.  —  To 

return  to  the  carriage  beam,  or  trimmer.  The  weight  to  be 
carried  upon  a  carriage  beam  is  compounded  of  two  loads  ; 
one  the  ordinary  or  distributed  load  upon  a  floor  beam,  as 


TOO        GIRDERS,    HEADERS   AND   CARRIAGE   BEAMS.      CHAP.  VII. 

shown  in  formula  (24)\  the  other  a  concentrated  load  from 
the  header.  Of  the  former  a  carriage  beam  is  required  to 
carry  one  half  as  much  as  an  ordinary  beam  ;  or,  the  load 
which  comes  upon  the  space  from  its  centre  half  way  to  the 
adjacent  common  beam.  This  is  the  half  of  that  shown  in 

formula  (2A),  or 

\acfl2  =  Bbd2 

The  symbol  W  in  formula  (23.)  represents  the  load  from 
the  header,  and  is  equal  (Art.  14-3)  to  \fng.  The  carriage 
beam  carries  half  this  load,  or  %fng  ;  hence 

\fng  =  W          or,  by  formula  (23.  \ 
mn  mn          ,   mn* 


Combining  this  with  the  formula  for  the  diffused  load,  we 
have 

\acfl2  +  afg  "^-  =  Bbd9  or 


This  is  a  rule  for  the  resistance  to  rupture  in  carriage 
beams  having  one  header.  (See  Art.  241,  and  caution  in 
Art.  88.) 

151.—  Example.  —  As  an  example,  let  it  be  required  to  show 
the  breadth  of  a  white  pine  carriage  beam  20  feet  long,  car- 
rying a  header  10  feet  long,  with  tail  beams  16  feet  long, 
in  a  floor  of  lo-inch  beams,  which  are  placed  15  inches 
from  centres  ;  and  where  the  load  per  superficial  foot  is  100 
pounds,  and  the  factor  of  safety  is  4. 

Transposing  formula  (29)  we  have 


_ 
b-af  ± 


CARRIAGE    BEAM  —  TWO    HEADERS.  IOI 

in  which  a  —  4,  f—  100,  c  =  15  inches  =  ij  feet,  /=  20, 
g  —  10,  n  —  16,  m  —  l—n  =  20—  16  =  4,  B  —  500  and 
d=  10.  Therefore, 

2  +  .ioxi6ax-A-) 
-  -  SLZ 


=  4  x  loo  x-—  —  -= 


5oox  10 


The  breadth  required  is  5.096,  or  sa}"  5  inches.  The 
trimmer  should  be  5  x  10  inches. 

152.— Carriage  Beam  with  Two'  Headers. — For  those 
cases  in  which  the  opening  in  the  floor  (Fig.  25)  occurs  at  or 
near  the  middle  (instead  of  being  at  one  side,  as  in  Fig.  22), 
two  headers  are  required  ;  consequently  the  carriage  beam, 
in  addition  to  the  load  upon  an  ordinary  beam,  has  to  carry 
tivo  concentrated  loads. 

To  obtain  a  rule  for  this  case  the  effect  produced  upon  a 
beam  by  two  concentrated  loads  will  first  be  considered. 

153.— Effect  of  Two  Weights  at  the  Location  of  One  of 
Them. — The  moment  of  one  weight  upon  a  beam  is  (Art.  56) 

W '-j~.    This  is  the  effect  at  the  point  of  location  of  the  weight. 

A  second  weight,  at  another  point,  will  produce  a  strain  at 
the  location  of  the  first  weight.  To  find  this  strain,  let  two 
weights,  W  and  V  (Fig.  24)  be  located  upon  a  beam  resting 
upon  two  supports,  A  and  B.  Let  the  distance  from  W  to 
the  support  which  may  be  reached  without  passing  the  other 
weight,  be  represented  by  ;;/,  and  the  distance  to  the  other 
support  by  ;/.  From  V  let  the  distances  to  the  supports 
be  designated  respectively  by  s  and  r  ;  s  and  n  being 
distances  from  the  same  support. 

The  letters  W  and  V,  representing  the  respective 
weights,  are  to  be  carefully  assigned  as  follows :  — Multiply 


IO2         GIRDERS,    HEADERS   AND   CARRIAGE   BEAMS.      CHAP.  VII. 

one  of  the  weights  by  its  distance  from  one  support,  and  the 
product  by  the  distance  from  the  other.  Treat  the  other 
weight  in  the  same  manner  ;  and  that  weight  which,  when 
so  multiplied,  shall  produce  the  greater  product  is  to  be 

called    W. 

For  example,  in  Fig.  24  let  the  two  weights  equal   8000  and 


w 


FIG.  24. 

6000,  /=  20,  the  distances  from  the  8000  weight  to  the  sup- 
ports equal  4  and  16,  and  those  from  the  6000  weight 
equal  5  and  15. 

Then  8000  x  4  x  16  —  5  12000 

and  6000  x  5  x  1  5  =  450000 

The  former  result  being  the  greater,  the  former  weight, 
Sooo,  is  to  be  called    W,   and  the  latter    V. 

The  moment  or  effect  of  the  weight    W    at  its  location  is 

equal,  as  before  stated,  to   W  -=-.      The  effect  of  the  weight 

V  at  the  point  W  will  (Art.  27)  be  equal  to  the  portion  of  V 
borne  at  A,  multiplied  by  the  arm  of  lever  m  (Arts.  34  and 
57).  The  portion  of  V  sustained  by  A  is  (Arts.  27  and  28), 

Vj  ;   hence  the  effect  of   V  at    W  will  be    Vj  x  m  =  V  ^. 
Adding  the  two  effects,  we  have 


This  is  the  total  effect  produced  at    W  by  the  two  weights. 


EFFECT  OF  TWO   CONCENTRATED   LOADS.  103 

In  like  manner  it  may  be  shown  that  the  'total  effect  at 
V  is 

/  / 

These  are  the  moments  or  total  effects  at  the  two  points 
of  location.  The  first,  when  modified  by  the  factor  of  safety 
a,  gives 

a  j  (Wn  +  Vs)  =  Sbd2  =  -bd* 

(see  Art.  35)  from  which  we  have 

4^(  Wn  +  Vs)  =  Bbd*  (30) 


for  the  dimensions  at    W.     Then,  also, 


4<t     (  Vr  +  Wm)  =  Bbd>  (31.) 


for  the  dimensions  at    V. 
(See  caution  in  Art.  88.) 


—  Example.  —  When  the  beam  is  to  be  of  equal  cross- 
section  throughout  its  length,  as  is  usually  the  case,  then 
formula  (30.),  giving  the  larger  of  the  two  results,  is  to  be  used. 

For  example,  let  a  weight  of  8000  pounds  be  placed  at 
3  feet  from  one  end  of  a  beam  12  feet  long  between  bearings, 
and  another  weight  of  3000  pounds  at  5  feet  from  the  other 
end. 

Then,  as  directed  in  Art.  153, 

8000  x  3  x  9  =  216000 
3000  x  5  x  7  =  105000 

The  weight  of  8000  pounds  having  given  the  larger  pro- 
duct, it  is  to  be  designated  by  W,  and  the  other  weight  by  V. 


104         GIRDERS,    HEADERS   AND    CARRIAGE   BEAMS.      CHAP.  VII. 

Making   a  •=.  4,    we  have  for  the  greater  effect  (form.  30.\ 

1M 

=  Bbd* 


4  x  4  x  —  x  ( 8ocx>  x  9  +  3000  x  5  )  =  Bbd~  =  348000 

and  with    £=$oo,      and     b=o-jd,    we  have 
B  x  o-jdx  d2  —  348000 

348000 

ds—-^^        -=004-29 
500  x  07 

d  =  9.98 

b  =  0-7  x  9-98  =  699 
or  the  beam  should  be    6.99x9.98,   or    7x10    inches. 

155.— Rule  for  Carriage  Beam  with  Two  Headers  and 
Two  Set§  of  Tail  Beam*. — Let  the  rules  of  Art.  153  be 
applied  to  the  case  of  a  carriage  beam  with  two  concentrated 
loads,  as  in  Fig.  25. 


FIG.  25. 

When  the  opening  in  the  floor  is  midway  between  the 
walls,  the  two  sets  of  tail  beams  are  of  equal  length  ;  or, 
m=s  ;  and  n=r  ;  therefore  mn=sr.  The  weights  are  also 
equal ;  therefore  Wmn  =  Vrs  ;  or,  the  strains  at  the  headers 


CARRIAGE   BEAM   WITH   TWO    HEADERS.  105 

are  equal.  By  moving  the  opening  from  the  middle,  the 
weight  at  the  header  carrying  the  longer  tail  beams  is  in- 
creased ;  so  also  the  product  of  the  distances  to  the  supports 
is  increased  ;  therefore  the  letter  W  is  to  be  put  at  that 
header  which  carries  the  longer  tail  beams,  for  then  the  pro- 
duct Wmn  will  exceed  the  product  Vrs. 

The  weight  at  W  is  equal  to  the  load  upon  one  end  of  the 
header  which  is  lodged  there  for  support.  This  is  equal  to 
(Arts.  14-3  and  150)  ^fgm  (  m  being  the  length  of  the  tail 
beams  sustained  by  this  header),  or  W=  \fgrn. 

In  like  manner  it  may  be  shown  that  V=  %fgs. 

By  substituting  these  values  of  W  and  V  in  formula 
(30.)  we  have 


In  addition  to  this  load,  the  carriage  beam  is  required  to 
carry  half  the  load  upon  a  common  beam,  or  half  that  shown 
at  formula  (24-),  or  \acfl*.  The  expression  for  the  full  effect 
at  W  therefore  is 


Bbd*  = 

Bbd2  =  af[m  (mn  +  s  ")  f  +  \cl*  ] 

In  like  manner  we  find  for  the  full  effect  at 

Bbd*  =  a/[s  (rs  +  m  *)  f 
(See  caution  in  Art.  88.) 


106        GIRDERS,    HEADERS    AND    CARRIAGE   BEAMS.      CHAP.  VII. 

These  two  formulas  (32.  and  33.)  give  the  sizes  of  the 
carriage  beam  at  W  and  V  respectively,  but  when  the 
beam  is  made  equal  in  size  throughout  its  length,  as  is 
usual,  the  larger  expression  (form.  32.)  is  to  be  used. 

156.— Example. — What  is  the  required  breadth  of  a 
Georgia  pine  carriage  beam  25  feet  long,  carrying  two 
headers  12  feet  long,  so  placed  as  to  provide  an  opening 
between  them  5  feet  wide;  the  tail  beams  being  15  feet 
long  on  one  side  of  the  opening  and  5  feet  long  on  the 
other  ;  the  floor  beams  being  14  inches  deep  and  placed  -18 
inches  from  centres  ;  the  load  per  superficial  foot  being  150 
pounds,  and  the  factor  of  safety  being  4? 

Taking  m  to  represent  the  longer  tail  beams,  we  have 
a  =  4,  7=150,  m=i$,  n  =  10,  J=5,  g  =  12,  /=  25, 
c  =  1 8  inches  =  i|  feet,  B  =  850  and  d  —  14. 

Formula  (32. \  now  becomes 


850x^x14'  =  4XI5of  I5(I5XIo+52)^7  +  iXIix252 


*=  'S+'  +       *  =  5-3* 


showing  that  the  breadth  should  be    5.38.    The  beam  may  be 
made    5^  x  14  inches. 

(57.—  Rule  for  Carriage  Beam  wills  Two  Headers  and 
One  Set  of  Tail  Beams.  —  The  preceding  discussion,  and  the 
rules  derived  therefrom,  are  applicable  to  cases  in  which  the 
two  headers  include  an  opening  between  them.  When  the 
headers  include  a  series  of  tail  beams  between  them,  leaving 
an  opening  at  each  wall  (Fig.  26),  then  the  loads  at  W  and  V 
are  equal  ;  for  the  total  load  is  that  which  is  upon  the  one 
series  of  tail  beams,  and  is  carried  in  equal  portions  at  the 
ends  of  the  two  headers  —  a  quarter  of  the  whole  load  at  each 


CARRIAGE  BEAM — TWO  HEADERS — ONE  SET  TAIL  BEAMS.    IO/ 


FIG.  26. 


end  of  each  header.  If  by  j  we  represent  the  length  of  the 
tail  beams,  we  have  W=  V=\jfg,  and  from  formula  (30.) 
we  have,  for  the  effect  at  Wy 


Add  to  this  half  the  load  upon  a  common  beam,    \acfl*    (Art. 
92),  and  we  have,  as  the  full  effect  at     W, 


and,  for  the  size  01  the  beam  at    W, 


=  Bbd*  ($4-) 


Similarly,  we  find  for  the  size  of  the  beam  at  V, 


af-~s  (r  +  m)  +  \cl  *     =  Bbd' 


108        GIRDERS,    HEADERS   AND   CARRIAGE   BEAMS.      CHAP.  VII. 

These  are  identical,  except  that  s  (r+  m)  in  (35.)  occupies 
the  place  of  m(n  +  s)  in  (34*)-  (See  caution  in  Art.  88.) 

As  in  Art.  I53,  care  must  be  taken  to  designate  by  the 
proper  symbols  the  weights  and  their  distances.  In  that 
article  the  proper  designation  was  found  by  putting  the  letter 
W  to  that  weight  which  when  multiplied  into  its  distances 
m  and  n  would  give  the  greater  product.  Here,  as  the 
weights  are  equal,  the  comparison  may  be  made  simply 
between  the  two  rectangles  mn  and  rs.  Of  these,  that 
will  give  the  greater  product  which  appertains  to  the 
weight  located  nearer  the  middle  of  the  beam ;  this  weight, 
therefore,  is  to  be  designated  by  Wy  and  will  be  found  at 
that  header  which  is  at  the  side  of  the  wider  opening.  The 
distances  m  and  n  appertain  to  the  weight  W.  The 
symbols  being  thus  carefully  arranged,  formula  (34-)  gives 
the  larger  result,  and  is  to  be  used  when  the  beam  is  to  be 
of  equal  sectional  area  throughout. 

I58.— Example. — To  show  the  application  of  this  rule,  let 
it  be  required  to  find  the  size  of  a  carriage  beam  in  a  tier  of 
beams  12  inches  deep  and  16  inches  from  centres,  with  a 
weight  per  superficial  foot  of  100  pounds.  In  this  case  what 
should  be  the  breadth  of  a  white  pine  carriage  beam  20  feet 
long  between  bearings,  carrying  two  headers  12  feet  long 
each,  with  one  series  of  tail  beams  10  feet  long  between  them, 
so  located  as  to  leave  an  opening  6  feet  wide  at  one  wall 
and  4  feet  at  the  other;  the  factor  of  safety  being  4  ? 
Here  we  have  the  two  distances  m  and  s  equal  to  6 
and  4,  and  putting  ;;/  for  the  larger  we  have  a  =  4, 
/=  100,  j—  10,  £-=12,  1=20,  m  =  6,  n  —  14,  5  =  4, 
c  =  i \,  .5=500  and  d=i2. 

Transposing  formula  (34)  to  find    b,    we  obtain 


CARRIAGE    BEAM — QUESTIONS.  109 

4X100      [~IOXI2 

b  =  -i      — =x  -x6(i4+4)  +  i><iVx202      —4-34 

500XI22     L      20  J 

The  breadth  is  required  to  be    4-34    inches,  and  the  size  of 
carnage  beam,  say    4|- x  12    inches.     (See  caution,  Art.  88.) 


QUESTIONS   FOR   PRACTICE. 


159. — A  building,  26  feet  wide  between  the  walls,  has  a 
tier  of  floor  beams  12  inches  deep  and  14  inches  from  cen- 
tres, supported  at  16  feet  from  one  of  the  walls  by  a  gir- 
der resting  upon  posts  set  15  feet  apart.  Upon  that  side  of 
the  building  where  the  girder  is  16  feet  distant  from  the  wall 
a  stair  opening  occurs,  extending  14  feet  along  the  wall,  and 
6  feet  wide.  The  floor  is  required  to  carry  150  pounds  per 
foot  superficial,  including  the  weights  of  the  materials  of  con- 
struction, with  a  factor  of  safety  of  4.  The  girder,  trim- 
mers and  header  all  to  be  of  Georgia  pine. 

NOTE. — The  resulting  answers  to  the  following  questions  will    be  smaller 
than  if  obtained  under  rules  in  Chapter  XVII.     (See  Art.  88.) 

160. — What  must  be  the  breadth  and  depth  of  the  girder, 
the  breadth  being  equal  to  55  hundredths  of  the  depth  ? 

161, — What  should  be  the  breadth  of  the  carriage  beams? 
162.— What  should  be  the  breadth  of  the  header? 

163. — What  should  be  the  area  of  cross-section  of  the 
bridle  iron  ? 


110          GIRDERS,    HEADERS   AND    CARRIAGE   BEAMS.     CHAP.  VII. 

164. — Another  opening  6  feet  wide  in  the  same  tier  of 
^eams,  has  headers  10  feet  long,  with  tail  beams  on  one  side 
6  feet  long  and  on  the  other  side  4  feet  long.  What 
should  be  the  breadth  of  the  carriage  beams  ? 

165. — What  ought  the  breadth  of  the  floor  beams  of  the 
aforesaid  floor  to  be  on  the  16  feet  side  of  the  girder,  if  of 
white  pine  ? 

166. — In  the  same  tier  of  beams  there  is  still  another  pair 
of  carriage  beams.  These  carry  two  headers  16  feet  long, 
and  the  two  headers  carry  between  them  one  series  of  tail 
beams  8  feet  long,  thus  forming  two  openings,  one  at  the 
girder  3  feet  wide  and  the  other  at  the  wall  5  feet  wide. 
What  should  be  the  breadth  of  these  carriage  beams  ? 


CHAPTER  VIII. 


GRAPHICAL    REPRESENTATIONS. 


ART.   167. — Advantages  of  Graphical  Representations. — 

In  the  discussion  of  the  subject  of  rupture  by  cross-strains, 
rules  have  been  given  by  which  the  effect  in  certain  cases  has 
been  ascertained  ;  for  example,  that  at  the  middle  of  a  beam 
which  rests  upon  two  supports ;  that  at  the  wall  in  the  case 
of  a  lever  inserted  in  the  wall ;  and  that  at  any  given  point 
in  the  length  of  a  beam  or  lever. 

These  rules  are  perhaps  sufficiently  manifest ;  but  when  it 
becomes  desirable  to  know  the  effect  of  the  load  in  a  new 
location,  or  under  other  change  of  conditions,  an  entirely 
new  computation  is  needed. 

To  obviate  the  necessity  for  this  labor,  and  to  fix  more 
strongly  upon  the  mind  the  rules  already  given,  the  method 
of  representing  strains  graphically,  or  by  diagrams,  is  useful, 
and  will  now  be  presented. 

168. — Strains  in  a  L.ever  measured  by  Scale. — In  Fig.  27 
we  have  a  lever  AB,  or  half  beam,  in 
which  the  destructive  energy  or  moment 
of  the  weight  P,  suspended  from  the 
free  end  B,  is  equal  to  the  product  of 
the  weight  into  the  arm  of  leverage  at 
the  end  of  which  it  acts  (Art.  34) ;  or 


FIG.  27. 

From   A    drop  the  vertical  line    AC  =  c,    make  it  by  any 
convenient  scale  equal  to   \IP,  and  join    C  and   B.     The  tri- 


112  GRAPHICAL    REPRESENTATIONS.  CHAP.  VIII. 

angle  ABC  forms  a  scale  upon  which  the  strain  produced  at 
any  point  in  AB  may  be  obtained,  simply  by  measurement ; 
for,  at  any  point,  D,  the  ordinate  DE  (—  y),  drawn  parallel 
with  the  line  AC,  is  equal  (measured  by  the  same  scale)  to 
the  strain  at  the  point  D.  In  the  two  homologous  triangles 
ABC  and  DBE,  we  have  this  proportion : 

I     7  CX 

*/:    c    ::    x   :  ,  == -^ 

By  construction    c  =  \IP,    therefore 

UPx 

y  =  --f  =  px 

equals  the  weight  into  the  arm  of  lever  at  the  end  of  which 
it  acts  ;  or  Px  =  y  is  the  destructive  energy  or  moment  of 
the  weight  P  at  the  point  D. 

In  this  equation  (Px  —  y)  since  P  is  constant,  the  value 
of  y  is  dependent  upon  that  of  x,  for  however  x  may  be 
varied,  y  will  vary  in  like  manner.  If  x  be  doubled,  y 
will  be  doubled  ;  if  x  be  multiplied  or  divided  by  any 
number,  y  will  require  to  be  multiplied  or  divided  by  the 
same  number. 

We  conclude  then  that  we  may  assign  any  value  to  x 
desirable,  or  select  any  point  in  AB  for  the  location  of  D, 
from  D  draw  an  ordinate  DE,  parallel  with  the  line  AC, 
and  measuring  the  ordinate  by  the  same  scale  by  which  c 
was  projected,  find  the  strain  or  destructive  energy  exerted 
upon  the  beam  at  the  selected  point  D. 

169. — Example— Rule  for  Dimensions. — For  example,  let 
P  —  100  and  /  =  20,  then  AB  —  \l  =  10,  and 

=    IO  X  IOO  =    IOOO 

Now  from  a  scale  of  equal  parts  (say  tenths  of  an  inch,  or 
any  other  convenient  dimensions),  lay  off  c  equal  to  ten  of 
the  divisions  of  the  scale  ;  then  each  division  represents  100 


CORRESPONDING   FORMULA.  113 


pounds  and  c=iooo  =  %/P.  Draw  the  line  CB,  and  from 
any  point  D  draw  the  ordinate  y.  Suppose  that  y, 
measured  by  the  same  scale,  is  found  to  equal  7^;  then 
the  strain  at  D  equals  J\  x  100  =  725  pounds. 

If  y  —  6,  then  the  strain  at  D  equals  600  pounds  ;  and 
so  of  any  other  ordinate,  its  measure  will  indicate  the  strain 
in  the  beam  at  the  end  of  that  ordinate. 

We  have,  therefore  [as  in  Art.  34,  formula  (#.)] 

Px  =  Sbd2 

and,  with  a  the  factor  of  safety,  and  putting  for  S  its 
equivalent  \B  (Art.  35), 


or,  4/ter  =  Bbd*  (36.) 

It  is  to  be  observed  that  the  b  and  d  of  this  formula 
are  those  required  at  D,  the  location  of  the  ordinate  y. 

When  x  equals  the  length  of  the  lever  AB,  equals  -J/, 
we  have 


2Pal  =  Bbd2 

and  if   P   be  taken  as  £  W,    W   being  the  load  at  the  centre 
of  a  whole  beam,  we  have 

2  *%Wal=Bbd* 

Wai  =  Bbd2 
the  same  as  formula 


170. — Graphical  Strains  in  a  Double  L,ever. — In    Fig.   28 

we    have    a  beam    AB    resting         . 

upon  a   point  at  the  middle    Cy 
and    carrying      the    two    equal       "Tx^      f 
loads  R  and  P  suspended  from       j[     \$  fi 
the  ends. 

The  half  of  this  beam,  or   CB, 
is  under  the  same  conditions  of  FlG<  *8' 

strain  as  the  beam  AB  in  Fig.  27,  and  since  the  weights  R  and 


GRAPHICAL   REPRESENTATIONS. 


CHAP.  VIII. 


P  are  equal,  and  C  is  at  the  middle  of  AB,  the  one  half  of 
the  beam,  or  AC,  is  strained  alike  with  the  other  half  CB. 
Therefore  a  strain  at  any  point  in  the  length  of  the  beam  is 
measured  by  an  ordinate  from  that  point  to  the  line  A  WB, 
and  formula  (36.)  is  applicable  to  this  case  also,  conditioned 
that  x  does  not  exceed  ^/. 

171. — Graphical  Strains  in  a  Beam. — In  Fig.  29  we  have  a 
beam   AB,  resting  upon  two  supports  A    and   B,  and  loaded 

at  middle  with  the  weight  W, 
one  half  of  which,  R,  is  borne 
upon  A,  and  the  other  half,  P, 
is  supported  by  B. 

This   beam    has    the    same 
strains  as  that  of  Fig.  28,  there- 
FlG-  29-  fore    (see   Art.    26)   the    same 

formula  (36.)  is  applicable,  namely  : 

tfax  =  Bbda 
P  =  ^W,    and  by  substitution 


(37.) 


2  Wax  =  Bbd3 


a  rule  applicable  to  this  case,  conditioned  that    x    shall  not 
exceed    £/. 

When    x  =    /    then  we  have 


Wai  =  Bbd2 

the  same  as  given  in  formula  (21.). 

Again,  if   x   be  diminished  until  it  shall  reach  zero,  then 

2  Wax  =  o 

or  the  strain  is  nothing.     This  is  evidently  correct,  as  the 
effect  of  the  weight,  in  producing  cross-strain,  disappears  at 


OF   THE    SHEARING   STRAIN. 


the  edge  of  the  bearing.  We  are  not  to  be  permitted,  how- 
ever, in  shaping  the  beam  to  its  exact  requirements,  to  re- 
move all  material  at  and  upon  the  bearing  wall,  for  there 
is  another  strain,  known  as  the  shearing  strain,  for  which 
provision  is  to  be  made  at  the  end  of  the  beam. 
This  strain  we  will  now  consider. 

(72. — Mature  of  the  Shearing  Strain. — The  nature  of  the 
shearing  strain,  as  well  as  of  the  cross-strain,  is  very  clearly 
shown  in  Fig.  30,  a  diagram  suggested  by  a  similar  one 
in  "Unwin's  Wrought-Iron  Bridges  and 
Roofs,  London,  1869." 

In  this  figure  a  semi-beam,  AB,  fixed 
in  a  wall  at  A,  is  cut  through  at  CD,  and 
the  severed  piece,  CB,  is  held  in  place 
by  means  of  a  strut  at  D  and  a  link  at 
C,  which  resist  the  compression  and  ten- 
sion due  to  the  cross-strain  arising  from 
the  weight  P ;  and  by  the  weight  R 
(equal  to  P  )  suspended  over  a  pulley  E, 


FIG.  30. 


which  prevents  the  severed  beam  from  sinking,  or  resists 
the  shearing  strain. 

As  the  link  C  and  strut  D  are  both  acting  in  a  horizon- 
tal direction,  they  can  have  no  effect  in  resisting  a  vertical 
strain,  consequently  the  weight  P  must  be  entirely  sustained 
by  the  counter-weight  R,  and  as  the  action  of  the  latter  is 
directly  opposite  to  that  of  the  former,  it  must  be  equal  to  it 
in  amount. 

In  the  above  arrangement  we  may  see  that  were  the 
strut  D  removed,  the  beam  CB,  under  the  action  of  the 
weight  Pj  would  revolve  upon  C  as  a  centre,  closing  the 
gap  at  the  bottom  ;  hence  the  strut  D  is  compressed. 

In   like   manner,  if  the   link   at    C    were   removed,  the 


Il6  GRAPHICAL    REPRESENTATIONS.  CHAP.  VIII. 

weight  P  would  cause  the  beam  to  revolve  on  D,  making 
wider  the  opening  at  the  top,  and  showing  that  the  link  C 
is  in  tension.  If  the  tension  at  C  be  represented  by  /,  the 
compression  at  D  by  c,  and  the  depth  CD  by  d,  then 

td  =  cd=Px  CB 

Disregarding  the  weight  of  the  beam,  the  shearing  strain 
at  CD  equals  the  weight  P.  As  this  strain  is  wholly  inde- 
pendent of  the  distance  between  C  and  B,  the  beam  may 
be  cut  at  any  point  in  its  length  with  a  like  result  as  to  the 
amount  of  the  shearing  strain.  At  every  point  we  shall 
have  R  =  P,  or  the  shearing  strain  equal  to  the  weight. 

If  the  weight  of  the  beam  be  included  in  the  considera- 
tion, the  shearing  strain  at  any  point  will  equal  the  weight 
P  plus  the  weight  of  so  much  of  the  beam  as  extends  beyond 
the  point  at  which  the  shearing  strain  is  considered. 

Let  CD  be  the  cross-section  at  which  it  is  required  to 
find  the  shearing  strain ;  let  JT  equal  the  distance  from  this 
cross-section  to  B,  in  feet ;  and  let  e  represent  the  weight 
per  foot  lineal  of  the  beam  ;  then  the  weight  of  the  piece  CB 
will  equal  ex,  and  the  shearing  strain  at  CD  will  equal 
P+  exy  or  the  destructive  energy  is 

D  ==  P  +  ex 

f73. — Transverse  and  Shearing  Strains  Compared. — Be- 
fore this  formula  can  be  available,  it  is  needed  to  know  the 
resistance  of  the  different  materials  to  this  kind  of  force. 
Experiments  have  been  made  upon  wrought-iron  which 
show  that  its  shearing  resistance  is  about  seventy-five  per 
cent  of  its  resistance  to  tension.  If,  in  the  absence  of  the  ex- 
periments necessary  to  establish  the  resistance  to  shearing 
in  materials  generally,  it  be  assumed  that  they  bear  the 


MEASURE   OF   SHEARING   STRAIN.  117 

same  proportion  to  their  tensile   resistance  as  is   found    in 
wrought-iron,  this  shearing  strength  may  be  put  equal  to 


in  which  T  equals  the  absolute  resistance  to  tension  per 
square  inch  of  cross-section. 

The  resistance  of  certain  woods  to  tension  may  be  found 
in  Table  XX. 

When   D  =  R   we  have 

P+ex=  \Tbd 

This  gives  bd,  or  the  area  of  cross-section,  equal  only  to  the 
destructive  energy.  In  this  case  rupture  would  ensue.  We 
therefore  introduce  the  factor  of  safety,  a,  and  have 

a(P+cx)=\TV&  (38} 

The  portion  of  T  considered  safe  is  from  one  sixth  to  one 
ninth.  We  then  have  a  —  6  to  a  =  9. 

As  an  example:  Suppose  a  semi-beam  (as  AB,  Fig.  30) 
of  white  pine  to  be  10  feet  long,  and  loaded  at  the  end  with 
P=  10,000  pounds  ;  what  would  be  the  required  area  of  cross- 
section  at  the  wall  ? 

Here  the  weight  of  the  beam  is  so  small  in  comparison 
with  the  load  P  that  it  may  be  neglected  in  the  computation. 
Throwing  it  out  of  the  formula,  we  have 


(39.} 
Let    a  —  9    and     T  •=•  12000  ;    then 

loooo  x  9  =  {.  x  1  2000  x  bd 

10000  x  o 

-?-—bd—iQ 

Jxl2OOO 


Il8  GRAPHICAL   REPRESENTATIONS.  CHAP.  VIII. 

To  compare  this  requirement   with   that  for  the    cross- 
strain,  we  make  use  of  the  formula  for  this  strain,  (19.), 

4Pan  =  Bbd* 
and,  making    a  =  4,  have 

4  x  10000  x  4  x  10  —  500  x  bda 

4  x  10000  x  4x  10        .  „ 

=  W  =  =  3200 


and,  making    d  =  16,  have 

b  x  i62  =  3200 


therefore  the  area  will  be     12^  x  16  =  200    square  inches. 

This  is  the  area  required  at  the  wall,  but  at  the  end  B, 
the  point  of  attachment  of  the  weight,  we  have  seen  (Fig. 
27)  that  the  destructive  energy  in  cross-strain  is  zero. 
Were  this  the  only  effect  produced  by  the  weight  P,  the 
beam  might  be  tapered  here  to  a  point.  Owing,  however, 
to  the  shearing  effect  of  the  weight,  we  find,  as  above,  a 
requirement  of  material  equal  to  10  inches  in  area,  or  the 
beam  12^  inches  wide  would  require  to  be  eight  tenths  of  an 
inch  thick  ;  and  the  rope  supporting  the  weight  should  be  so 
attached  as  to  have  a  bearing  across  the  whole  width  of  the 
piece. 

174.—  Rule    for  Shearing    Strain    at    Ends    of   Beams.— 

The  shearing  strains  at  the  two  supports  upon  which  a  beam 
is  laid  are  together  equal  to  the  weight  of  the  beam  and  the 
load  laid  upon  it.  If  the  beam  be  of  equal  cross-section 
throughout  its  length,  and  the  load  upon  the  beam  be  located 
at  the  middle,  or  symmetrically  about  the  middle,  then  the 


SIZE   OF   BEAM   AT   ENDS.  IIQ 

weight  of  the  beam  and  its  load  will  be  sustained  half  upon 
each  support.  In  this  case,  the  shearing  strain  at  the  two 
supports  will  be  equal,  and  each  equal  to  half  the  total  load. 
Putting  W  for  the  load  upon  the  beam,  and  el  for  the  weight 
of  the  beam,  then  for  the  shearing  strain  at  each  end  of  the 
beam  we  have 


Putting  this  equal  to  the  safe  resistance  [see  formula  (38,  ,), 
Art.  173]  we  shall  have 


(W  +  el)  =  \Tbd 

bd  (40.) 


When  the  load  is  not  at  the  middle  nor  symmetrically 
disposed  about  the  middle,  the  portion  borne  upon  each 
support  may  be  found  by  formulas  (3.)  and  (4>),  Art.  27.  The 
shearing  strain  at  each  support  is  equal  to  the  reaction  of 
the  support  or  to  the  load  it  bears. 

175.—  Resistance  tio  Side  Pressure.  —  Beyond  the  fore- 
going considerations,  there  is  still  another  of  some  impor- 
tance. Care  should  be  taken  that  the  surfaces  of  contact  of 
the  wall  and  the  beam  are  of  sufficient  area  to  be  unyielding. 
Usually  the  wall  composed  of  brick  or  stone  is  so  firm  that 
there  need  be  no  apprehension  of  its  failure,  and  yet  it  is 
well  to  know  that  it  is  safe.  It  should,  therefore,  be  carefully 
considered,  to  see  that  the  given  surface  is  sufficiently  large 
for  the  given  material  to  carry  safely  the  weight  proposed  to 
be  distributed  over  it.  In  calculations  for  heavy  roof  trusses 
this  precaution  is  particularly  necessary. 

The  upper  surface  of  the  joint,  or  underside  of  the  beam, 


120  GRAPHICAL    REPRESENTATIONS.  CHAP.  VIII. 

requires. especial  attention.  This  is  usually  of  timber,  and 
parallel  with  the  fibres  of  the  material.  The  pressure  upon 
the  surface  tends  to  compress  these  fibres  more  compact!)7 
together  by  closing  the  cells  or  pores  which  occur  between 
the  fibres.  When  pressed  in  this  way,  timber  is  much  more 
easily  crushed,  as  may  readily  be  supposed,  than  when  the 
pressure  is  applied  at  the  ends  of  the  fibres  in  a  line  parallel 
with  their  direction. 

The  resistance  to  side  pressure  approaches  the  resist- 
ance to  end  pressure  in  proportion  to  the  hardness  of  the 
material. 

By  experiments  made  by  the  author  some  years  since,  to 
test  the  side  resistance,  results  of  which  are  recorded  in  the 
American  House  Carpenter,  page  179,  it  appears  that  the  hard- 
est woods,  such  as  lignum-vitae  and  live  oak,  will  resist  about 
i£  times  the  pressure  endwise  that  they  will  sidewise ;  ash, 
if  times  ;  St.  Domingo  mahogany,  twice;  Baywood  mahog- 
any, oak,  maple  and  hickory,  about  3  times  ;  locust,  black 
walnut,  cherry  and  white  oak,  about  3^  times  ;  Georgia  pine, 
Ohio  pine  and  whitewood,  about  4  times  ;  chestnut,  5  times  ; 
spruce  and  white  pine,  8  times  ;  and  hemlock,  9  times.  Their 
resistance  to  side  pressure  is  in  proportion  to  the  solidity  of 
the  material,  or  inversely  in  proportion  to  the  size  of  the 
pores  of  the  wood. 

In  the  above  classification,  the  comparison  is  not  that  of 
the  absolute  resistance  of  the  several  kinds  of  wood  to  side 
pressure.  It  is  only  a  comparison  of  the  results  of  the  two 
pressures  on  the  same  wood.  Whitewood,  classed  above 
with  Georgia  pine,  resists  sidewise  only  as  much,  absolutely, 
as  white  pine.  Its  power  of  resistance  to  end  pressure  is  the 
lowest  of  any  of  the  woods,  being  but  one  half  that  of  white 
pine. 

The  average  effectual  resistance  to  side  pressure  per 
square  inch  of  surface,  /,  for 


BREADTH   OF   BEARING  ON   WALLS.  121 

Spruce  =250  pounds. 

White  pine  =  300        " 

Hemlock  =  300        " 

Whitewood  =  300        " 

Georgia  pine  —  850 

Oak  =  950 

Under  these  pressures  only  a  slight  impression  is  made, 
and  the  woods  may  be  safely  trusted  with  these  respective 
amounts. 

176.  —  Bearing  Surface  of  Beams  upon  Walls.  —  The  sur- 
face of  ,  the  beam  in  contact  with  the  wall  must  be  sufficient 
in  extent  to  insure  that  it  shall  not  be  exposed  to  more  pres- 
sure than  is  above  shown  to  be  safe.  If  b  equal  the  breadth 
of  the  beam,  h  the  length  of  the  bearing  surface,  and  p  the 
resistance  per  inch,  as  above,  then  the  total  resistance  equals 

R  =  bhp 

The  destructive  energy  for  one  end  of  the  beam  is,  as 
before  (Art.  174), 

D  = 


When  there  is  equilibrium,  then   R  =  D,  or 

l)  =  bhp 


Owing  to  the  deflection  of  the  beam  by  the  load  upon  it, 
its  extreme  ends  may  be  slightly  raised  from  off  the  bearing 
surface,  and  in  consequence  the  pressure  be  concentrated  at 
the  edge  of  the  wall.  No  serious  effect  will  ensue  from  this, 
for  if  the  pressure  be  greater  than  the  timber  can  resist  at 
the  edge,  the  fibres  will  be  crushed  there,  but  only  suffi- 
ciently so  to  allow  the  surface  of  contact  to  extend  towards 


122  GRAPHICAL   REPRESENTATIONS.  CHAP.  VIII. 

the  end  of  the  beam,  until  it  is  so  enlarged  as  to  effectually 
resist  any  further  crushing. 

Beams  which  are  likely  to  be  depressed  considerably 
should  have  their  ends  formed  so  that  their  under  surface 
will  coincide  throughout  with  the  wall  surface  when  the 
greatest  load  shall  have  been  put  upon  them. 

177.— Example  to  Find  Bearing  Surface.  —  Let  a  white 
pine  carriage  beam  6  inches  wide,  24  feet  long  between 
bearings,  and  weighing  15  pounds  per  lineal  foot,  be  loaded 
with  12,000  pounds,  equally  distributed  over  its  length. 
What  should  be  the  length  of  the  bearing  upon  each  wall  ? 

By  transposition,  formula  (41.)  becomes 

W+el  =  k 
2bp 

In  this  case,  W  —  12,000,  e  —  15,  /  =  24,  b  —  6,  and 
p  =  300;  then 

12000  +  15x24        , 

?  =  h  =  3.43 
2x6x300 

or  the  end  of  the  beam  must  extend  upon  the  wall,  say  3^ 
inches.  The  usual  bearing  for  floor  beams,  which  is  4 
inches,  would  in  this  case  be  amply  sufficient. 

Where  the  concentrated  weight  is  so  large  in  comparison 
with  the  weight  of  the  beam,  the  latter  Aveight  may  be  neg- 
lected without  any  serious  result ;  for  had  we  considered  the 
12,000  pounds  only,  in  the  above  example,  the  value  of  h 
would  have  been  3.33,  only  a  tenth  of  an  inch  shorter  than 
the  former  result. 

178.— Shape  of  Side  of  Beam,  Graphically  Expres§ed.— 

As  will  be  observed,  we  have  digressed  from  the  principal 
subject.  This  became  necessary  in  order  to  explain  the 
apparently  anomalous  result  of  leaving  the  beam  without  any 


SHAPE   OF   SIDE    OF   LEVER.  123 

support  at  the  ends.  For  it  was  seen  that  in  an  application 
of  the  formula  for  cross-strains  the  requirement  of  material 
gradually  lessened  towards  the  ends  of  the  beam,  until  at 
the  very  edge  of  the  bearings  it  entirely  disappeared. 

To  prevent  the  beam,  with  its  load,  from  falling  as  a  dead 
weight  between  the  bearings  ;  or,  to  provide  against  the 
shearing  strain,  as  well  as  against  the  crushing  of  the  material 
upon  its  bearings,  we  have  turned  aside  so  far  as  seemed  to 
be  needed.  And  before  returning  to  the  main  subject,  it  may 
be  well  here  to  show  that  the  line  CB  in  Figs.  27  and  29,  limit- 
ing the  ordinates  of  cross-strain  in  the  lever  and  beam,  does 
not  show,  as  might  be  supposed,  the  shape  of  the  depth  of  a 
lever  or  beam  having  a  cross-section  of  equal  strength 
throughout  its  length.  A  short  consideration  of  the  relation 
between  the  strains  at  given  points  in  the  length  will  show 
the  true  shape. 

By  construction,  c,  Fig.  27,  is  equal  to  %tP,  and  from  this 
we  have  shown  (Art.  168)  that 


and  when  the  destructive   energy  and  the   resistance  are 
equal 

\lP^Sbd2  and 

Px  =  Sbdf  from  which 

c\y\\  Sbd*  :  Sbdf  and  when 

S  and   b  are  constant 

c  :  y  :  :  d3  :  df 

or,  the  ordinates  are   in  proportion   to  the  squares  of  the 
depths,  and  not  directly  as  the  depths  themselves. 

From  these  ordinates,  however,  the  shape  of  the  side  of 
the  lever  may  be  directly  found  by  taking  their  square  roots. 
For  let  AB  in  Fig.  3*  be  the  upper  edge  of  the  lever,  and 


124 


GRAPHICAL   REPRESENTATIONS. 


CHAP.  VIII. 


T    t 


CB  the  line  limiting  the  ordinates  of  cross  strain.     Then,  if 

AD  be  made  equal  to  the  square 
root  of  AC,  and,  corresponding- 
ly, dt,  din  dilit  etc.,  be  each  made 
respectively  equal  to  the  square 
root  of  the  ordinate  upon  which 
it  lies,  and  if  a  line  be  drawn 
through  the  ends  of  dn  d:t,  dllt, 
etc.,  this  line,  DEB,  will  limit  the 
shape  of  the  lever. 

This  curve  line  is  a  semi-para- 
bola, with  its  vertex  at  B  and 
its  base  vertical  at  AD.  By  con- 
struction, each  ordinate  y  is  in  proportion  to  x,  its  dis- 
tance from  B,  or  (since  y  equals  d2)  d2  is  in  proportion  to 
x,  a  property  of  the  parabola.  Hence  to  obtain  the  shape  of 
the  lower  edge  of  the  lever,  any  method  of  describing  a  para- 
bola may  be  used,  making  AD,  its  base,  equal  to  (form.  19.) 


FIG.  31. 


FIG.  32. 


Bb 

As  a  whole  beam  is  in  like 
condition  with  two  semi-beams, 
as  to  the  cross  strains,  there- 
fore the  shape  of  a  whole  beam 
of  equal  strength  throughout 
its  length  is  that  given  by  two 
semi-parabolas  placed  base  to 
base,  as  in  Fig.  32. 


QUESTIONS   FOR   PRACTICE. 


(79. — In  a  semi-beam,  or  lever,  10  feet  long,  fixed  in  a 
wall,  and  loaded  at  the  free  end  with  3672  pounds,  what  is 
the  destructive  energy  at  the  wall  ? 

180. — Make  a  graphic  representation  of  the  above  by  a 
horizontal  scale  of  one  foot  to  the  inch,  and  a  vertical  scale 
of  1000  foot-pounds  to  the  inch.  What  is  the  height  CA  of 
the  triangle  of  cross-strains,  in  terms  of  the  scale  selected  ? 

(81. — Measuring  horizontal  distances  from  the  free  end, 
what  are  the  lengths,  by  the  scale,  of  the  respective  ordinates 
at  the  several  distances  of  5,  6,  7,  8  and  9  feet;  and  what 
the  amount  of  cross-strain  corresponding  thereto  at  these 
several  points  in  the  beam  ? 

182. — What  will  be  the  required  depth  at  the  wall,  and  at 
9  and  8  feet  respectively  from  the  free  end  ;  the  lever  being 
of  Georgia  pine,  6  inches  broad,  and  the  factor  of  safety  4? 

183. — In  a  white  pine  beam,  4  inches  broad,  16  feet  long 
between  bearings,  and  loaded  at  the  middle  with  3250 
pounds,  what  should  be  the  respective  depths  at  the  several 
distances  of  3,  5,  7  and  8  feet  from  one  end,  the  factor  of 
safety  being  4? 

184. — A  white  pine  semi-beam,  12  feet  long  and  4  inches 
broad,  is  loaded  with  693  pounds  at  the  free  end,  including 
the  effect  of  the  weight  of  the  beam  itself.  The  factor  of 
safety  is  4,  the  beam  is  of  constant  breadth  and  depth 


126  GRAPHICAL   REPRESENTATIONS.  CHAP.  VIII. 

throughout  its  length,  and  its  weight  is   30   pounds  per  cubic 
foot. 

What  is  its  required  depth  at  the  wall  ? 

What  is  the  weight  suspended  from  the  end  of  the  beam  ? 

What  is  the  shearing  strain  at  the  wall  ? 

What  is  the  shearing  strain  at   5    feet  from  the  wall  ? 

185. — A  beam  of  Georgia  pine,  4  inches  broad  and  20  feet 
long,  is  loaded  at  the  middle  with  9644!  pounds.  The  beam 
is  17  inches  high  at  the  middle,  and  tapered  in  parabolic 
curves  to  each  end.  The  material  of  the  beam  is  estimated 
at  48  pounds  per  cubic  foot.  What  is  the  weight  of  the 
beam? 

186. — What  is  the  shearing  strain  at  each  wall  ? 
With  a   factor   of  safety   of   9,   how    high   is   the   beam 
required  to  be  at  the  ends  to  resist  the  shearing  strain  safely  ? 

(87. — How  far  upon  each  wall  is  the  beam  required  to 
extend,  in  order  to  prevent  crushing  of  the  material  ? 


CHAPTER   IX. 

STRAINS   REPRESENTED   GRAPHICALLY. 

ART.  188. — Graphic  Method  Extended  to  Other  Cases. — 

In  Figs.  27,  28  and  29,  with  a  given  maximum  strain  upon  a 
semi-beam,  or  upon  a  full  beam,  we  have  a  ready  method  of 
finding  the  strain  at  any  given  point  in  the  length. 

This  simple  method  of  ascertaining  the  strain  at  any 
point,  graphically,  is  based  upon  a  principle  which  is  applic- 
able to  strained  beams  under  conditions  other  than  those 
given,  as  will  now  be  shown. 

189. — Application  to  Double  Lever  with  Unequal  Arms. — 

In  Figs.  28  and  29  the  load  upon  the  beam  is  at  the  middle. 
But  it  may  be  shown  that  the  triangle  of  strains  is  applicable 
in  cases  where  the  load  is  not  at  the  middle. 

Let  R  and    P,  Fig.  33,   represent  two  unequal  weights, 


FIG.  33. 

suspended  from  the  ends  of  a  balanced  lever    AB.     From 
the  law  of  the  lever,  we  have  (Art.  27) 


Rm  —  Pn 


123 


STRAINS   REPRESENTED   GRAPHICALLY.       CHAP.  IX. 


If  CD,  called  g,  be  made  of  a  length  to  represent  Pn,  then 
will  it  also  represent  Rm  ;  for  Rm  =  Pn.  Hence,  since  the 
triangle  BCD  is  the  triangle  of  strains,  in  which  an  ordinate, 
y,  showing  the  strain  at  any  given  point  in  DB,  may  be 
drawn,  therefore  the  triangle  ACD  will  give  ordinates,  y' , 
measuring  the  strains  at  the  points  in  AD,  from  which  they 
may  be  drawn  ;  or,  since 

Pn  :  g  : :  Px  :  y 


>.«£* 

n 


so  also 


Rm  :  g  ::  Rx'   :  / 


(4$-) 


f90. — Application  to  Beam  with  Weight  at  Any  Point.— 

In  Fig.  34,   AB    represents  a  beam  supported  at  each  end, 
carrying  a  load    W  at  a  point  nearer  to   A    than  to   B.     This 


w 


FIG.  34. 

beam  is  strained  in  all  respects  like  that  in  Fig.  33,  except 
that  the  strains  are  in  reversed  order.  Therefore  an  ordi- 
nate, y,  drawn  across  the  triangle  BC W,  will  indicate  the 
strain  at  the  point  of  its  location.  So  an  ordinate,  /,  across 
the  triangle  ACW,  will  indicate  the  strain  at  its  point  of 


SCALE   OF   STRAINS — WEIGHT  AT  ANY   POINT.  1 29 

location.  Or, generally,  the  two  triangles  ACW  and  BCW 
limit  the  ordinates  \vhich  measure  the  strains  at  any  point  in 
the  length  of  the  beam.  Thus  when 

g  =.  Pn  =  Rm  we  have 

y  —  Px      and      /  =  Rx' 

and  since  P=  W™       and       R  =  Wj      (Art.  27) 

we  have  y  =  W  ~,  x  (44-) 

y<=WUjx>  (45) 

Now,  since  Rm  —  Pn  =g,  equals  the  destructive  energy  of 
the  weight  at  its  location,  therefore  any  ordinate  across  the 
triangles  ACW  and  BCW  equals,  when  measured  by  the 
same  scale,  the  destructive  energy  at  the  location  of  that 
ordinate,  and  when  the  resistance  is  equal  to  the  destructive 
energy  we  have  for  the  strain  at  any  point  to  the  right  of 
the  weight 


Putting  for    5   its  equivalent  \B  (Arts.  35  and  57)  to  agree 
with  the  unit  of  dimensions,  we  have,  for  the  safe  weight, 


(46.) 


which,  with   x  at  its  maximum  equal  to    n,  is  identical  with 
formula  (23). 

For  the  safe  weight  at  any  point  to  the  left  of  the  weight 
we  have 

4Wajx'=zBbd*  (47.) 

191.—  Example.—  As   an   example    in   the   application   of 
these   expressions,  let  it  be  required  to  find  the  strains  at 


130 


STRAINS   REPRESENTED    GRAPHICALLY.       CHAP.  IX. 


various  points  in  the  length  of  a  white  pine  beam,  the 
maximum  strain  being  given. 

Let  the  beam  be  10  feet  long  and  loaded  with  2000 
pounds  at  a  point  three  feet  from  the  left-hand  end. 

What  is  the  strain  at  the  location  of  the  weight?  What 
are  the  several  strains  at  2,  4  and  6  feet  from  the  right- 
hand  end  and  at  2  feet  from  the  left-hand  end  ? 

Take  first  the  strains  to  the  right. 


.m 


Here,  by  formula  (44-),   y  =  W  jx,   and  with    x   at   its 
maximum  we  have 


y  —  2000  x  —  x  7  =  4200 


In  Fig.  35,  make  the  length  between  the  bearings  A  and 
B  by  any  scale,  equal  to  10  feet,  and  CW,  or  g,  equal  to 
42  units  of  any  other  scale.  Then  each  of  these  units  will 


\     s 
\ 


I    >--"- 


FIG.  35. 


represent  100  pounds  of  strain.  The  number  of  units  in 
the  length  of  the  ordinates,  y,  at  the  several  distances,  x 
equal  to  2,  4  and  6  feet,  and  of  x'  —  2  feet,  will  give,  when 
multiplied  by  TOO,  the  strains  at  these  several  points. 
Thus  it  will  be  found  that, 


DEPTH   OF   BEAM  —  WEIGHT  AT  ANY   POINT.  131 

at   2  feet  from  B,    y  —  12,  and  12  x  100  —  1200; 

"    4  "       "       B,    y  =  24,       "  24  x  100  —  2400  ; 

"   6  "       "       B,   y  —  36,       "  36  x  loo  =  3600  ; 

and   "    2  "       "      A,  y'  —  28,       "  28  x  100  =  2800. 

Now,  if  it  be  required  to  find  the  proper  depth  of  the 
beam  at  these  several  points,  we  take,  for  the  right-hand 
end,  formula 


in  which  W  represents  2000  pounds,  the  weight  upon 
the  beam,  and  in  which  W-j-x  will  give  the  strain  at  each 
ordinate  ;  and  by  transposition  have 


and  if    a  =  4,    B  —  500    and  b  =  3,     we  have 


500  x  3  x  10 


=  6 


and  therefore 

when        x  —  2     then     ^2  =  6-4x2=J2-8     and  ^=3.58 

4:^4       "        d2  =  6-4x4^25-6       "  ^=5.06 

^r  =  6       "        ^'  =  6.4x6  —  38-4       "  ^=6.20 

"    x=n  =  ?       "        d*  —  6-4  x  7  =44-8       "  ^=6-69 

For  the  left-hand  end  we  use  formula  (47-) 


132  STRAINS   REPRESENTED   GRAPHICALLY.       CHAP.  IX. 

4  X  2000  X  4  X  7 

d2  =  --  — ^x'—iA.-^x1 

500  x  3  x  10 

and  hence, 

when  x'  =  2    then    */'  =  14-93  x  2  =  29-9  and  </—  5-47 

"     y  =  »i=3       "       ;/*:=:  14-93  x  3  =44.8      "     d  =  6-6g 

This  last  result  agrees  with  the  last  from  the  right-hand 
end,  as  it  should,  for  they  are  both  for  the  same  location. 
The  above  results  are  all  obtained  by  computations,  but  the 
value  of  d*,  at  as  many  points  as  may  be  desired,  can  be 
obtained  by  scale,  in  a  similar  way  with  the  ordinates  for  the 
destructive  energy  ;  but  this  scale,  for  the  purpose  of  obtain- 
ing the  depths,  must  be  made  with  the  principal  ordinate,  gt 
equal  to  the  requirement 


(see  form.  #$.),  and  then  the  square  root  of  each  ordinate 
drawn  across  the  scale  will  be  the  required  depth  at  its 
location. 

For  example :  Make  g,  by  any  convenient  scale,  equal 
to  44-8  as  above  required  ;  then  the  several  values  of  d*  at 
2,  4  and  6  feet  may  be  found  by  measuring  the  ordinates 
drawn  at  these  several  distances  from  B. 

The  square  root  of  each  ordinate  will  equal  the  depth  of 
the  beam  there.  The  results  obtained  by  measurements, 
although  not  exact  to  the  last  decimal,  are  yet  sufficiently 
exact  for  all  practical  purposes.  If  it  be  required  to  find  the 
exact  dimension,  this  may  be  done  by  computation,  as  shown, 
and  the  diagram  will  then  serve  the  very  useful  purpose  of 
checking  the  result  against  any  serious  error  in  the  calcula- 
tion. 


MEASURE   OF   STRAIN   FROM   TWO   WEIGHTS. 


133 


192.— Graphical  Strains  toy  Two  Weiglit§. — The  value  of 
graphic  representations  is  manifest  where  two  or  more 
weights  are  carried  at  as  many  points  upon  a  beam. 

In  Fig,  36  we  have  a  beam  carrying  two  weights  A'  and  Br. 


The  destructive  energy  of  the  weight  A',  at  its  location, 
is  equal  to  (Art.  56) 


and  the  destructive  energy  of  the  weight   B'  ',   at  its  location, 
is  equal  to 

D"  =  B'  ~ 


mn 


Make  AE  equal  to  A'-j-   by  any  convenient  scale.     By  the 

TS 

same  scale  make  BF  equal  to  B'-j-.     Draw  the  lines  CE  and 

DE,   CF  and  DF. 

Now,  while  AE  represents  the  effect  of  the  weight  A1  at 
the  point  A,  so  also  AG  measures  (A rt.  190)  the  effect,  at  the 
same  point,  of  the  weight  B'  ;  therefore  make  EJ  equal  to 
AG,  then  AJ  is  the  total  effect  at  A  of  both  weights. 


134 


STRAINS   REPRESENTED   GRAPHICALLY.       CHAP.  IX. 


In  like  manner  (FK  being  made  equal  to  BH),  BK 
measures  the  total  effect  at  B.  Draw  the  line  CJKD.  and 
by  dropping  a  vertical  ordinate  from  any  point  in  the  beam 
CD  to  this  line,  we  have  the  total  strain  in  the  beam  at 
that  point. 

193.— Demonstration. — The  above  may  be  proved,  as 
follows : 

First.  Let  the  ordinate  occur  between  the  two  weights 
as  LM,  Fig.  37. 

Extend  the  lines    CF,  DE  and    JK,  till  they  meet  at    R 
and   5,  and  draw   CR  and  DS. 


FIG.  37. 


Now  the  effect  of  B'  at  B,  is  measured  by  BF,  and  at  L 
by  LP  (Art.\B9).  Also  the  effect  of  A'  at  A,  is  measured 
by  AE,  and  at  L  by  LN.  The  joint  effect  of  A'  and  B' 
at  L,  is  thus  LP+LN,  and  if  it  can  be  shown  that  PM 
equals  LN,  then 

LP+LN=LP+PM  =  LM 

equals  the  joint  effect  of  the  two  weights  A'  and  £',  at  L. 

In  two  triangles  of  equal  base  and  altitude,  two  lines 
drawn  parallel  to  the  respective  bases,  and  at  equal  alti- 
tudes, are  equal;  from  which,  conversely,  if  two  triangles  of 
equal  base  have  equal  lines  drawn  parallel  to  the  base,  and 


SCALE   OF   STRAINS  —  DEMONSTRATION.  135 

at  equal  altitudes,  then  the  altitudes  of  the  two  triangles 
are  equal.  In  the  present  case  we  have  AE  =  GJ\  for 
A  G  —  EJ  by  construction  ;  and  if,  to  each  of  these  equals 
we  add  the  common  quantity  GE,  the  sums  will  be 
equal,  or 


AE=  GJ 

The  two  triangles    ADE  and    GSJ  are  therefore  standing 
upon  equal  bases,  AE  and    GJ. 

Moreover,  at  equal  distances,  AB,  from  the  line  of  bases 
AJ,  and  parallel  with  it,  we  have  the  two  lines  BH  and 
FK,  made  equal  by  construction.  Consequently,  the  two 
triangles  have  equal  altitudes.  Hence  all  lines  drawn  across 
them,  parallel  with  and  at  equal  distances  from  the  base,  are 
equal,  and  therefore  LN  and  PM,  having  these  properties, 
are  equal,  and  LM=LP+LN  equals  the  true  measure  of 
the  strain  induced  at  L  by  the  weights  A'  and  B'  ;  or,  in 
general,  any  vertical  ordinate  drawn  across  AJKB  will 
measure  the  total  strain  caused  by  the  two  weights  at  the 
location  of  the  ordinate. 


—  Damonsf  ration—  Rule    for   the    Varying    Depths.— 

Second.  Let  the  ordinate  occur  at  one  end,  between  B  and 
D,  as  OQ,  Fig.  37. 

Here  we  have  OT  for  the  strain  caused  by  A',  and 
O  V  for  the  strain  caused  by  B'  ;  or  the  total  strain  equals 
OT+OV. 

Now  if  VQ  can  be  proved  equal  to   OT,  we  shall  have 


equal  to  the  total  strain  at  O. 

We  have  the  two  triangles    BDH  and  FDK,  with  bases 


136 


STRAINS   REPRESENTED    GRAPHICALLY.       CHAP.  IX. 


W 


BH  and  FK,  made  equal  by  construction,  and  with  equal 
altitudes  BD,  and  we  have  the  two  lines  OT  and  VQ 
drawn  parallel  with,  and  at  equal  altitudes  ( BO )  from  the 
base;  consequently  OT  and  VQ  are  equal,  and  OQ  meas- 
ures the  total  strain  of  the  two  weights  at  O\  or,  in  gene- 
ral, any  vertical  ordinate  drawn  across  BDK  will  measure 
the  total  strain  at  the  location  of  the  ordinate. 

Since  it  may  be  shown  in  like  manner  that  any  vertical 
ordinate  drawn  across  ACJ  will  measure  the  total  strain  at 
its  location,  therefore  we  conclude  that  a  vertical  ordinate 
from  any  point  in  the  beam  CD  to  the  line  CJKD  will 
show  the  total  strain  in  the  beam  at  that  point. 

In  practice,  the  scale  of  strains  CJKD  may  be  con- 
structed as  just  shown,  in  detail,  but  more  directly  by 
obtaining  the  points  J  and  K  in  the  following  manner  : 

We  have  for  the  joint  effect  of  the  two  weights  at  the 
location  of  one  of  them,  A,  (see  Art.  153) 


which  becomes,  on  changing   W  and    V  into  A1  and  Bf, 

**'"\'+ffs)  (51.) 


equals  the  length  ol  the  ordinate  AJ. 


TWO   WEIGHTS — STRAINS   AND   DEPTHS.  137 

In  like  manner  we  have 


(52.) 


for  the  length  of  the  line  BK. 

The  points  J  and  K  are  to  be  obtained  by  these  expres- 
sions. The  scale  is  then  completed  by  connecting  these 
points  and  the  ends  of  the  beam  by  the  line  CJKD.  The 
strain  at  any  point  in  the  beam  may  then  be  readily  meas- 
ured, sufficiently  near  for  all  practical  purposes. 

If,  however,  the  exact  strain  is  desired,  this  may  be 
obtained  as  follows  : 

Putting  g  for  AJ,  p  for  BK,  and  //  for  AB,  we  have 
for  the  several  ordinates 

s  :  p  ::  x  :  y 

y=tx  (53.) 

S 

m  :  g  :  :  x'  :  yr 


h  •  p-g  ::.*":  y"  -g 
h(y"-g)  =  x»(p-g) 
hy"-hg  =  x"(p-g) 
hy"  =    x"(p—g)  +  hg 


If  it  be  required  to  know  the  depth  of  the  beam  at  every 
point,  to  accord  with  the  strain  there,  then,  instead  of  mak- 
ing the  two  principal  ordinates  as  above  shown,  find  their 
lengths  thus  : 


138  STRAINS   REPRESENTED   GRAPHICALLY.       CHAP.  IX. 

By  formulas  (30.)  and  (51.)  make  AJ  equal  to 


m 


d*  = 


(56.) 


Bb 


and  by  formulas  (31)  and  (52.)  make    BK  equal  to 


4a-(B'r  +  Am) 


(57.) 


Draw  the  line  CJKD,  and  then  an  ordinate  drawn 
across  this  scale  at  any  point  will  give  the  square  of  the  depth 
at  that  point.  The  square  root  of  this  length  will  be  the 
required  depth  there. 

195. — Graphical  Strains  by  Three  Weights. — In  Fig.  38 
we  have  a  graphical  representation  of  the  strains  resulting 
from  three  weights. 


FIG.  38. 

This  figure  is  constructed  by  making  AJ  equal  to  the 
moment  of  A'  at  A,  BK  equal  to  the  moment  of  B'  at 
B,  and  CL  equal  to  the  moment  of  C'  at  C,  all  by  the  same 


MEASURE   OF   STRAINS   FROM   THREE   WEIGHTS.  139 

scale.  Connect  J,  K  and  L  each  with  the  ends  of  the 
beam  E  and  D.  Make  JF  equal  to  AM  +  AN,  KG 
equal  to  BO  +  BP,  and  LH  equal  to  CQ  +  CR. 

Join  E,  F,  G,  H  and  D,  and  this  line  will  be  the  boun- 
dary of  any  vertical  ordinate  from  any  point  in  ED,  which, 
by  the  same  scale  as  used  for  AJ,  etc.,  will  measure  the 
strain  at  the  location  of  the  ordinate. 

In  this  diagram,  the  points  F,  G  and  H  may  be  found 
directly,  as  follows  : 

To  find  F,  we  have  (Art.  153)   A'~  for  the  effect  of  A', 

B'—J-  for  Bf,  and  so,  in  like  manner,  we  may  have  C'  ~j- 
for  that  of  C'.  Added  together,  these  will  equal 

AF  =  ™(A'n  +  B's  +  C'v  )  (58.) 

To  find  Gt  we  have  A'^  for  A',  B'--  for  Bf,  and 
C-r  for  C  ;  which  together  give 

n^       A'ms  +  B'rs  +  C'rv 

±>(JT    =    — 


To  find  H,  we  have  Af~  for  A',  B'~  for  Bf,  and  C~ 
for  C  ;  which  added,  will  equal 

CH  —  -(A'm  +  Bfr  - 

If  it  be  desirable,  the  strains  may,  as  in  the  last  figure,  be 
computed  ;  for  putting  g  for  AF,  p  for  BG,  k  for  CH, 
h  for  AB,  and  q  for  BC,  we  have,  for  an  ordinate  between 
C  and  D, 


140 


STRAINS   REPRESENTED    GRAPHICALLY.       CHAP.  IX. 


FIG.  38. 

v  :  k  :  :  x  :  y 

9*&  ^ 

For  an  ordinate  between  E   and  A  we  have 

m  :  g  :  :  x'  :  y' 

m 
For  an  ordinate  between  A  and  B  we  have,  as  in  Fig.  37, 


y"  = 


h 


(63) 


and  for  ordinates  occurring  between  B  and  C  we  have 

y  =  tJLx">  +  k  (64.) 

These  expressions  give  the  strains  at  any  point,  due  to  the 
three  weights. 

In  like  manner,  we  may  find  the  strain  at  any  point  in  a 
beam,  arising  from  any  number  of  weights. 

To  obtain  the  squares  of  the  depths  at  various  points  by 
scale,  make  AF  equal  to 


THREE   EQUAL  WEIGHTS   SYMMETRICALLY   DISPOSED.    141 


Make  BG  equal  to 

A'ms'+  B'rs  +  Crv 


Make  CH  equal  to 


4a^(A'm  +  B'r  +  Ct) 

/  fay  \ 

d2  = : (67-) 

Bb 


The  square  roots  of  ordinates  upon  this  scale  will  give 
the  depths  required  at  their  several  locations. 

196.  —  Graphical  Strains  by  Three  Equal  Weights  Equa- 
bly Disposed.  —  Let  us  now  consider  the  effect  of  equal 
weights,  equably  disposed. 

In  Fig.  39  we  have  three  equal  weights,  L,  placed  at  equal 
distances  apart  upon  a  beam,  ED,  the  distance  from  either 
wall  to  its  nearest  weight  being  one  half  that  between  any 
two  of  the  weights  ;  or, 


EA  -  CD  = 


The  line  EFGHD  is  obtained  as  directed  for  Fig.  38.  It 
may  also  be  obtained  analytically,  thus  : 

First.  The  line  AF,  or  the  effect  at  A  of  the  three 
weights,  equals  the  sum  of  the  three  lines  AJ,  AO  and 

AN. 


142 


STRAINS  REPRESENTED  GRAPHICALLY.   CHAP.  IX. 


FIG.  39. 

Let  EA  =  CD  =  t,   and    AD  —  h,    then   t  +  h  =  /,    and 
(Art.  56) 


tx/t 


th 


as  per  Art.  195. 


CDxEA  _       fr//x*        t     /A 
-L--J--  -L     t-  -^L  t 


or 


th 


.  th 


Second.  The  line  BG,  or  the  effect  at  B  of  the  three 
weights,  is  equal  to  the  sum  of  the  line  BK  and  twice  the 
line  BQ. 

Let    EB  =  /,    BD  —  h,  and   /  +  //  =  /;  then 


T 
BK  =  Z-- 


th 


and 


FOUR  EQUAL  WEIGHTS   SYMMETRICALLY   DISPOSED.     143 


Third.  The  effect  at  C  produced  by  the  three  weights  is 
equal  to  that  at  A. 
We  have,  then, 


for  the  total  effect  at  A,        AF  =    £ 
B, 


tt  tt 


th 
I 
th 


197. — Graphical  Strains  by  Four  Equal  Weights  Equably 
Disposed. — When  there  are  four  equal  weights,  as  in  Fig.  40, 
similarly  disposed  as  in  Fig.  39,  the  effect  at  A  is, 


\ 


'M 


-.-v---—  .. 


0 


I 


FIG.  40. 


from  load  at  A, 


hxt 

L~     ~~ 


ht 


C, 
"       'JD, 


ht 

~T 
ht 


144  STRAINS   REPRESENTED    GRAPHICALLY.      CHAP.  IX. 

or  the  total  effect  at  A,  of  the  four  weights,  is 


The  effect  at  B  is, 

from  load  at  A,  L^-j^  =  \L-r 


C, 


or  the  total  effect  at  Z?,  of  the  four  weights,  is 


The  effect  at  C  is  equal  to  that  at  B,  and  the  effect  at 
D  is  equal  to  that  at  A. 

198.  —  Graphical  Strains  by  Five  Equal  Weights  Equably 
Disposed.—  When  there  are  five  equal  weights,  as  in  Fig.  41, 
similarly  disposed  as  those  in  Fig.  39,  the  effect  at  A  is, 

Thxt  T  ht 

from  load  at  A,  L—j-       —  %L-j 

B,  ^hxt-^lL  — 

11          C,  |A:X-/7.=  |£y 


"      M, 


FIVE   EQUAL   WEIGHTS   SYMMETRICALLY  DISPOSED.      145 

L  L  L  L  L 


ABC 


M 


or  the  total  effect  at  A,  of  all  the  weights,  is 


The  total  effect  at  B  is, 


from  load  at  A, 


c. 


II  tt 


"       M, 


ht 
-j- 

ht 
-l- 

y 
-f 


or  the  total  effect  at  B,  of  all  the  weights,  is 


146  STRAINS   REPRESENTED   GRAPHICALLY.      CHAP.  IX. 

• 


The  total  effect  at  C  is, 


from  load  at  A,  ty  x  h-  —  \L^ 

/  / 


,  = 

or  the  total  effect  at  C,  of  all  the  weights,  is 


7 


The  effects  produced  at  D  and  J/  are,  respectively,  like 
those  at  B  and  A. 


199. — General  Result*  from  Equal  Weights  Equably  Dis- 
posed.—In  looking  over  the  results  here  obtained,  it  will  be 

seen  that  in  each  case  the  effect  is  equal  to  gL—r,  in  which 

g  is  put  for  the  numerical  coefficient,  L  for  any  one  of  the 
equal  weights  with  which  the  beam  is  loaded,  /  and  h  the 
respective  distances  from  the  point  at  which  the  strain  is 
being  measured  to  the  ends  of  the  beam,  and  /  for  the  length 
of  the  beam.  All  of  these  are  simple  quantities  except  the 
coefficient  g,  and  this  it  will  be  shown  is  subject  to  a  certain 
law  and  may  be  stated  in  general  terms. 


TOTAL   STRAIN   AT   LOCATION   OF   FIRST   WEIGHT.         147 

2  00  a  —  General  Expression  for  Full  Strain  at  First  Weight. 

—  The  coefficient  g  is  a  fraction,  having  its  numerator  and 
denominator  both  dependent  upon  the  number  of  weights 
upon  the  beam. 

Let  us  first  consider  the  value  of  the  numerator  in 
measuring  the  effect  of  the  weights  at  A,  the  location  of  the 
first  weight  from  the  left. 

With  three  weights,  g,  the  coefficient,  was  |  +  f  +  |  =  f, 
the  numerators  being  1  +  3  +  5=9. 

With  four  weights,   g  was  equal  to    •  -  =  —  ,  the 

numerators  being    1+3  +  5  +  7=16. 

I  I  ^  I  £  I  7  I  Q         25 

With    five    weights,    g    was   equal   to  -  =  —  , 

and  the  numerators    1+3  +  5  +  7  +  9  =  25. 

In  general,  we  shall  find  that  the  numerator  of  the  frac- 
tion g,  is  in  all  cases  equal  to  the  sum  of  an  arithmetical 
progression  comprising  the  odd  numbers  i,  3,  5,  etc.,  to  n 
terms  ;  ;/  being  put  to  represent  the  number  of  weights 
upon  the  beam,  the  first  term  being  unity,  and  the  last  being 
2n—\. 

To  find  the  sum  of  this  progression,  we  have 


in  which   S  =  the  sum,   a  —  the  first  term,  /  =  the  last  term, 
and    n  =  the  number  of  terms  ;  or 


_  I  +  (2n—  i)n       n  +  2n2—n 

O  —  > — r^ n:  ft' 


Hence,  the  numerator  of  the  coefficient  of  the  expression 
showing  the  effect  of  any  number  of  weights  at  the  location, 
Ay  of  the  first  weight,  is  equal  to  the  square  of  the  number 
of  weights  ;  thus,  when  there  are 


148  STRAINS   REPRESENTED    GRAPHICALLY.      CHAP.  IX. 

2  weights,  n  =  2,  and  the  numerator  =  22  =    4 

3  "  « =  3,  "                "            =  32  -    9 

4  »  =  4»  "  "            =  42  =  16 

5  "  «=5.  "              "           =  52  =  25 

6  0  =  6,  "  "            =  62  =  36 

and  so  for  any  number  of  weights. 

In  considering  the  value  of  the  denominator  of  g  it  will  be 
observed  that  it  is  derived  by  taking  the  value  of  h  in  each 
case  in  terms  of  /.  With  three  weights,  //  =  5^  ;  with  four 
weights,  //  =  7/  ;  and  with  five  weights,  //  =  qt ;  so  that  in 
general,  h-=-(2n—\)t.  The  denominator  of  the  fraction 
generally,  therefore,  is  20—1. 

na 
The  value  of  the  coefficient  is,  consequently,   —    — ,  and 

the  full  effect  at    A    of  any  number  of  equal  weights  equably 

« '      _  /// 

disposed  upon  a  beam  is  -  L  —.-  . 

2n—\       I 

201.— General  Expression  for  Full  Strain  at  Second 
Weight — For  the  effect  at  the  location  B  we  have  the  ex- 
pression pL—.-_ ;  in  which  the  same  quantities  occur  as  before, 

except  in  the  case  of  the  coefficient  /. 

This  coefficient  is  composed  of  two  classes  of  fractions. 
The  first  of  these  is  based  upon  the  relation  between  the  dis- 
tances EA  and  EB,  and  since  EA  is  in  all  cases  equal  to  -J- 
of  EB,  therefore  this  part  of  the  coefficient  /  will  be  equal 
to  i 

In  the  second  fraction  of  the  coefficient,  the  numerator  is, 
as  in  the  case  at  A,  equal  to  the  sum  of  an  arithmetical  pro- 
gression, but  extending  one  less  in  the  number  of  the  terms, 
so  that  in  place  of  ns  we  put  (n—  i)2. 

The  denominator  is  found  by  taking  ;/— i  for  «,  or 
2(0—1)—!,  equal  to  20—3,  for  20—1.  The  value  of 


TOTAL   STRAIN   AT   LOCATION   OF   SECOND   WEIGHT.      149 

this  fraction  is  therefore   -  —  -~  .      To  this,  adding  the  first 
fraction,  we  have 


2/2-3 
and  for  the  full  effect  at  B,  of  all  the  weights, 

(t+  -<!=%* 
V      z«—  3/    / 

From  the  above,  the  value  of  the  coefficient  /  is  as  follows 

(2  —  if 
with  2  weights,  /  =  i  +  ^-x2)_3=i  +  -f     =| 

'    3        "  ,  =  t  +        =-  =  i  +  i    =V 


"    5 


The  numerators  of  these  results  are  in  the  order  of  2n, 
$n,  Sn,  nn  and  14;?;  the  numerals  differing  by  3.  The  de- 
nominators are  the  products  of  i,  3,  5,  7  and  9,  each  by  3. 
We  may  continue  therefore  the  values  to  any  number  of 
weights  by  following  these  laws,  thus 

f  •    i  17  X  7         IIQ 

for  7  weights,  /  = 


for  8  weights,  /  = 


HX3        33 

20  x  8  __  160 
i3><3"~    ~39~ 


or,  in  general,  the  effect  at  B  for  any  number  of  weights  may 
be  had  directly  from  the  previous  expression. 


150  STRAINS   REPRESENTED   GRAPHICALLY.      CHAP.   IX. 

202.— General     Expression     for     Full     Strain     at     Any 
Weight. — For  the  sum  of  effects  at  C,  it  is  seen  that  we  have 

kL-r-,  and  it  can  be  shown  that  the  coefficient  k  is  the  sum 

in 2\a 

of  two  fractions — namely,  f  and  -      _  or 


For  the  effect  at  D  we  have 


For  the  effect  at  E  we  have 


or,  putting  them  in  sequence,  we  have 

(n-o}' 


at      A      the  effect     g=    f 


C        "        "         k= 
D        "        "         u= 

E        ,         ., 


n- 


GENERAL  EXPRESSION— TOTAL  STRAIN  AT  ANY  WEIGHT.    15 1 

and  so  for  any  number  of  weights  upon  one  end  of  the 
beam. 

An  examination  of  this  series  shows  that  in  the  first  of 
the  two  fractions  the  numerator  is  equal  to  the  square  of  the 
number  of  weights  preceding  the  one  under  consideration  ; 
for  instance,  at  A,  where  there  are  no  weights  preceding, 
we  have  the  numerator  o ;  at  B  there  is  one  weight  preced- 
ing, and  hence  the  numerator  is  i2  equals  i  ;  at  C  there  are 
two  weights  preceding,  hence  the  numerator  equals  22  equals 
4  ;  at  D  there  are  three  weights,  hence  the  numerator  equals 
32  equals  9  ;  etc.  For  the  denominator  of  the  first  fraction  we 
have,  for  the  several  cases  in  consecutive  order,  the  values 
J>  3>  5>  7>  etc. ;  an  arithmetical  series  of  the  odd  numbers. 

In  the  second  fraction  we  have  a  numerator  equal  to  the 
square  of  the  difference  between  n  and  the  number  of  weights 
preceding  the  one  at  which  the  strain  is  being  measured ; 
and  a  denominator  of  2n  minus  the  denominator  of  the  first 
fraction. 

Let  r  represent  in  any  case  the  number  of  weights  pre- 
ceding the  one  at  the  location  of  which  we  wish  to  know 
the  strain.  Then  we  shall  have,  as  the  coefficient  of  the  effect 
at  that  point, 

r*  (n-r)a 

and  for  the  full  effect,  or  the  destructive  energy, 

D  =  L--(  ^—  +       <*-£-.    \  (68.) 

I  \  2r+  i       2n  —  (2r+i)  / 

in  which  L  represents  one  of  the  equal  weights  with  which 
the  beam  is  loaded  ;  //  the  distance  from  the  weight  at  which 
the  strain  in  the  beam  is  being  measured  to  the  right-hand 
end  of  the  beam  ;  t  the  distance  from  the  same  point  to  the 
left-hand  end  ;  /  =  h  •+  /  the  length  of  the  beam  between  sup- 


152  STRAINS    REPRESENTED    GRAPHICALLY.      CHAP.  IX. 

ports  ;  n  the  number  of  equal  weights  equally  disposed  upon 
the  beam,  as  in  Fig.  41  ;  and  r  the  number  of  weights  between 
the  point  where  the  strain  is  measured  and  the  left-hand  end 
of  the  beam,  not  including  the  one  at  the  point  where  the 
strain  is  measured. 

203.—  Example.—  What  is  the  strain  at  the  fifth  weight 
from  the  left-hand  end  of  a  beam  22  feet  long,  loaded  with  1  1 
weights  of  100  pounds  each  ;  the  weights  placed  at  equal 
distances  from  centres,  and  the  distance  from  each  end  of  the 
beam  to  the  centre  of  the  nearest  weight  being  equal  to  half 
the  distance  between  the  centres  of  any  two  adjoining 
weights?  Here  the  distance  between  centres  of  weights 
will  be  2  feet,  t  will  equal  9  feet,  and  h  will  equal  13 
feet,  L  =  100,  n  —  n,  and  r  —  4. 

From  these  the  strain  at  the  fifth  weight  will  be  (form. 
68.) 


D  =  ioox-  +  --     =  2950 

22    V  8+1       22  —  (8+  1) 


QUESTIONS    FOR    PRACTICE. 


204-.  — A  beam  12  feet  long  is  loaded  at  4  feet  from  the 
left-hand  end  with  4000  pounds.  What  is  the  strain  at  that 
point  ? 

205. — What  are  the  strains,  respectively,  at  2,  4,  and  6 
feet  from  the  right-hand  end  ? 

206. — A  beam  14  feet  long  is  loaded  with  two  weights; 
one,  A',  weighing  3000  pounds,  is  located  at  4  feet  from  the 
left-hand  end  ;  the  other,  B' ,  weighing  5000  pounds,  is  at  6 
feet  from  the  right-hand  end. 

What  strain  is  caused  by  these  two  weights  at  the 
point  A  ? 

What  strain  is  caused  at  Bl 

207. — In  the  above  beam  what  strain  is  caused  by  the 
two  weights  at  a  point  2  feet  from  the  left-hand  end  ? 

What  strain  is  caused  at  a  point  2  feet  from  the  right- 
hand  end? 

What  strain  is  produced  at  the  middle  of  the  beam  ? 

208. — Abeam  20  feet  long  is  loaded  with  three  weights; 
one,  A',  of  3000  pounds,  at  3  feet  from  the  left-hand  end; 
one,  B'9  of  2000  pounds,  at  1 1  feet  from  the  same  end  ;  and 
the  third  weight,  C' ,  of  4000  pounds,  at  4  feet  from  the 
right-hand  end. 


154         STRAINS  REPRESENTED  GRAPHICALLY.   CHAP.  IX. 

What  is  the  full  effect  of  the  three  weights  at  the  location 
of  each  weight,  at  2  feet  from  the  left-hand  end,  at  2  feet 
from  the  right-hand  end,  at  6  feet  from  the  same  end,  and  at 
the  middle  of  the  beam  ? 

209. — Abeam  16  feet  long  is  loaded  with  20  weights  of 
zoo  pounds  each,  the  weights  being  equally  distributed. 

What  strain  do  these  weights  produce  in  the  beam  at  the 
ninth  weight  from  one  end  ? 


CHAPTER  X. 

STRAINS   FROM    UNIFORMLY    DISTRIBUTED   LOADS. 

ART.  210. — Extinction  Between  a  Series  of  Concentrated 
Weights  and  a  Thoroughly  Distributed  ILoad. — The  distribu- 
tion of  the  load  upon  a  beam,  as  shown  in  Figs.  39,  40  and  41, 
is  essentially  that  of  a  uniform  distribution  over  the  entire 
length  of  the  beam.  For  if  the  beam  be  divided  into  as 
many  parts  as  there  are  weights,  by  vertical  lines  located 
midway  between  each  two  weights,  it  is  seen  that  the  parts 
into  which  these  lines  divide  the  beam  are  all  equal  one 
with  another,  and  the  weight  upon  each  part  is  located  'in  a 
vertical  line  passing  through  the  centre  of  gravity  of  that 
part.  Hence  this  beam,  taken  with  the  loads  upon  it,  is  an 
apparently  parallel  case  with  a  beam  having  an  equally 
distributed  load. 

An  application  of  formula  (68.),  however,  will  show  that 
the  case  is  that  of  a  beam  loaded  with  a  series  of  concentrated 
weights,  and  not  with  a  thoroughly  distributed  load,  although 
it  closely  approximates  the  latter.  We  find  that  the  results 
of  computations  made  with  this  formula  differ  according  to 
the  number  of  weights  upon  the  beam,  but  approach  a  cer- 
tain limit  as  the  number  of  weights  is  increased  ;  a  limit 
which  is  that  of  a  beam  with  an  equally  distributed  load. 

211. — Demonstration. — For  example,  let  us  find  by  for- 
mula (68.)  the  effects  at  the  middle  of  the  beam  under 
differing  numbers  of  weights. 


1  56   STRAINS  FROM  UNIFORMLY  DISTRIBUTED  LOADS.   CHAP.  X. 

We    may  modify    the    formula    to    suit    this    case,    for 
Lxn  =  U,    when    U   equals    the    total    weight    upon    the 

beam,  or   L  =  —  ,   and    h  =  t  =  \l. 

By  substituting  these  values,  we  have 


2r+ 


(69.) 


To  apply  this  modified  formula  to  the  question  : 

First.  Let    there   be   five   weights   equally  disposed,   or 
n  =  5  ;     then    r  —  2,   and  we  have 


Second.  Let  there  be  nine  weights  or    n  —  9,    then    r  —  4, 
and  we  have 

*=~ 


If  «  =  25,  then  r  —  12,  and 


Fourth.  \i  n—  101,  then  r  =  50,  and 

+  W)  = 


SERIES  OF  CONCENTRATED  LOADS.  157 

Comparing  the  coefficients  of  these  several  results,  we 
have 

when    n—      5,  the  coefficient  =  |~f        =J-fTV 

"          «=        9>  "  " 

"          »=     25,  «  " 

"       »  =  101,  "  "          =  -AWr  =  4  + 

The  result  in  all  cases  is  equal  to  a  half,  plus  a  fraction 
which  decreases  as  n  increases,  or  which  has  unity  for  its 
numerator,  and  a  denominator  equal  to  twice  the  square  of  n. 

The  coefficient  may  be  expressed  then  by   \  +  — 

Now,  when  the  number  of  weights  is  unlimited,  or  the 
load  thoroughly  and  equally  distributed  over  the  whole 
length,  then  n  is  infinite,  and  the  denominator  of  the  last 
fraction  becomes  infinity.  In  this  case,  the  fraction  itself 
equals  zero  and  consequently  vanishes. 

Hence  the  coefficient  tends  towards  •£,  and  with  the  loads 
subdivided  to  the  last  degree,  and  infinite  in  number,  actual- 
ly becomes  \  ;  for,  with  these  conditions  fulfilled  the  case 
is  actually  that  of  an  equally  distributed  load,  and  then 

x  =  \U-  =  i*7/.  (See  Art.  59.) 

This  value  of  the  coefficient  may  be  concisely  derived 
by  the  use  of  the  calculus,  as  will  now  be  shown. 


212. — Demonstration  by  the  Calculus. — To  obtain  a  for- 
mula to  represent  the  strain  caused  at  any  point  by  an  equally 
distributed  load,  let  RPTS,  Fig.  42,  represent  graphically 
an  equally  distributed  load,  SR  being  equal  to  TP,  and 
let  it  be  required  to  find  the  ordinate  EF,  equal  to  the 
effect  at  any  point  E,  caused  by  the  whole  load. 


158    STRAINS  FROM  UNIFORMLY  DISTRIBUTED  LOADS.   CHAP.  X. 

F     D 
''  C 


C              B                               P 
A    » 


I 


FIG.  42. 

To  do  this  we  may  proceed  as  follows  :  Let  the  ordinate 
AG  represent  by  scale  the  strain  caused  at  A  by  a  small 
weight  A',  concentrated  at  A.  Then  will  EJ  represent 
(Art.  190)  the  effect  of  A'  at  E.  Again,  let  the  ordinate 
BH  represent  by  scale  the  strain  at  B  caused  by  a  small 
weight  B' ,  concentrated  at  B.  Then  will  EK  represent 
the  effect  of  B'  at  E.  The  sum  of  these,  EJ+EK,  will 
equal  the  joint  effect  of  the  weights  A'  and  B'  at  E.  Or 
(Art.  190) 

A'hx     B'tx, 
U~    'I   '          I 


Let  the  loads  A'  and  B'  be  very  small ;  equal  to  a  small 
portion  of  the  equally  distributed  load  SRPT,  and  repre- 
sented graphically  by  the  thin  vertical  slices  at  A  and  B 
respectively,  and  let  these  slices  be  reduced  to  the  smallest 
possible  thickness.  By  the  rules  of  the  calculus  we  may 
represent  the  thickness  of  the  slices,  when  infinitely  reduced, 
by  dx,  the  differential  of  x,  or  rate  of  increase.  If  e  be  put 
to  represent  the  weight  per  lineal  foot  of  the  equally  dis- 
tributed load  SRPT,  then  edx  will  represent  the  weight 
of  the  thin  slice  at  A,  or  equal  A'.  So  also  edxt  will 
represent  the  weight  of  the  slice  at  B,  or  equal  B'. 


STRAINS   COMPUTED   BY   THE   CALCULUS.  159 

Substituting  these  values  for    A'  and  B'    in  the  above 
expression,  we  obtain 


~       ehxdx      etx.dx.        e  ,  ,     ,  , 

D  —  — - — +  — '- — '-  =  j  (hxdx  +  txpX 


This  is  the  effect  at  E  of  the  two  loads  at  A  and  B,  but 
these  loads  are  infinitesimally  small,  therefore  the  expression 
is  to  be  considered  merely  as  the  sum  of  the  differential,  or 
rates  of  increase  of  the  strains  produced  by  the  two  parts 
into  which  the  whole  of  the  equally  distributed  load  RSTP 
is  divided  by  the  ordinate  EF.  The  strain  itself  is  to  be 
had  by  the  integral  which  is  to  be  derived  from  the  above 
differential  of  the  strain.  Therefore,  by  integration,  we  have 
(Arts.  4-62  and  463) 

^  ( hxdx  +  txtdxt )  =  4  (\hx*  +  \tx?}  —  y 


By  integrating  between  x  —  o  and  x  =  /,  also  between 
xt  •=.  o  and  xt  —  /i,  or  making  the  integral  definite,  we 
have 


but  h  =  I  -  t 

therefore  ht  =  (I  —  t)  t 

and  h2  =  (l-t)a 
therefore 


=  (l-t)t+(l-t}s  =  lt—t*+l*—2lt+t*  =  Is -It 


l6o   STRAINS  FROM  UNIFORMLY  DISTRIBUTED  LOADS.    CHAP.  X. 

and  the  formula 

et 
y  =  —  j  (ht  +  ks)  becomes 

*=Ti(l'-K> 

y  =  \et(l-f)  (70.) 

This  result  gives  the  value  of  the  ordinate  y,  drawn  at 
any  point,  and  is  comparable  with  the  formula  for  the  para- 
bola*, in  which  /  equals  the  base,  and  the  maximum  ordi- 
nate, y,  equals  the  height.  Therefore,  if  the  curve  line 
RFDP  be  that  of  the  parabola,  it  will  limit  all  the  ordi- 
nates,  y,  which  may  be  drawn  from  the  line  RP. 

In  the  above  discussion  e  was  put  for  the  weight  of  one 
foot  lineal  of  the  load,  therefore  the  whole  load  U  equals 

el,   or  e  =  —  .     If  in   formula   (70.)  we  substitute  for  e  this 
value  of  it,  we  have 


and  when    h  —  t  —  \l  we  have,   for    the    ordinate    at    its 
maximum  or  at  the  centre, 


y  = 

(72.} 


*  For  here  we  have  an  ordinate  to  the  curve  from  any  point  in  the  base,  which 
is  in  proportion  to  the  rectangle  [t  x  (/ — /)]  of  the  two  parts  into  which  the 
base  is  divided  by  that  point,  a  property  of  the  parabola.  (See  Cape's  Mathe- 
matics, 1850,  Vol.  II.,  p.  48.) 


COMPARISON   OF   RESULTS.  l6l 

We  thus  see  that  the  true  value  of  the  coefficient 
discussed  in  Art.  211  is  equal  to  one  half. 

This  result  (£67)  is  the  effect  at  the  middle  of  the  beam, 
and  shows  that  an  equally  distributed  load  will  need  to  be 
twice  the  weight  of  a  concentrated  load  to  produce  like 
effects  upon  any  given  beam  ;  a  like  result  with  that  which 
was  obtained  in  another  way  at  Art.  59. 

213. — Distinction  Shown   by  Scales  of  Strains. — By    the 

calculus,  the  coefficient,  as  has  just  been  shown,  is  equal  to|, 
but  those  by  formula  (69.)  exceed  i  by  a  certain  fraction 
(Art.  211). 

A  comparison  of  the  scales  of  strains  in  'Figs.  41  and  42 
will  show  that  the  line  limiting  the  ordinates  is  not  a  para- 
bola, but  a  polygonal  line.  In  proportion  to  the  increase  in 
the  number  of  the  weights,  and  their  consequent  diminution 
in  size  and  distance  apart,  this  polygonal  figure  approximates 
the  parabolic  curve  ;  and  in  like  proportion  do  the  corre- 
sponding coefficients  approach  the  coefficient  obtained  by 
the  calculus;  until  finally,  when  the  number  of  the  weights 
becomes  infinite,  or  the  load  is  absolutely  an  equably  distrib- 
uted one,  then  the  coefficients  are  identical.  The  difference 
between  the  two  expressions  is  that  which  is  shown  between 
the  areas  of  the  polygonal  and  parabolic  figures. 

214. — Effect  at  Any  Point  by  an  Equally  Distributed 
Load. — One  other  lesson  may  be  learned  from  this  discus- 
sion. 

It  has  been  shown  (Arts.  59  and  61)  that  the  effect  at  the 
middle  of  the  beam,  from  an  equably  distributed  weight,  is, 
equal  to  that  which  would  be  produced  by  just  one  half  of 
the  weight  if  concentrated  there  ;  and  now  we  see  (Arts.  2(1 
and  212)  that  this  proportion  holds  good,  not  only  at  the 
middle  of  the  beam,  but  also  at  any  point  in  its  length. 


162    STRAINS  FROM  UNIFORMLY  DISTRIBUTED  LOADS.   CHAP.  X. 

The  expression  (71.)  just  obtained, 


gives  the  effect  produced  by  an  equally  distributed  load  at 
any  point  in  the  beam. 

It  was  shown  (Art.  56)  that  the  effect  at  any  point  of  a 
load  concentrated  at  that  point,  is  equal  to 

W™-  wht 
I  ~l 

Now  when  the  effects  in  the  two  cases  are  equal,  we  have 


or,  4*7  =  W  ±,i 

showing  that  when  the  effects  at  any  point  are  equal,  the 
concentrated  load  is  equal  to  just  half  of  the  uniformly 
distributed  load. 


215.—  Shape  of  Side  of  Beam  for  an  Equably  Distributed 
Load.  —  We  have  seen  (form.  71.)  that  the  effect  at  any  point 
in  a  beam  from  an  equably  distributed  load  is 


and  that  the  curve  drawn  through  the  ends  of  a  series  of 
ordinates  obtained  by  this  formula  is  a  parabola  (Art.  212, 
foot  note). 

From  this  may  easily  be  derived  the  form  of  the  depth  of 
a  beam  (the  breadth  being  constant),  which  shall  be  equally 
strong  throughout  its  length  to  bear  safely  an  equably  dis- 


ECONOMIC   FORM   OF   BEAM.  163 

tributed  load.     The  formula  (71.)   gives  the  strain  at   any 
point,  and  when  put  equal  to  the  resistance  (Art.  35)  is 

-  Sbd* 


r> 

Substituting  for    5    its   value   —   we   have  for  the  safe 
weight  (Art.  73) 


f  ,  .  ,  „       2UaAt 

from  which  a*         ^^ 


This  gives  the  square  of  the  depth  at  any  point,  and  when 
h  —  t  =    l    we  have 


equals  the  square  of  the  depth  at  the  middle. 

Now   make     CD,   Fig.  43,  equal  by  formula  (73.}   to    d* 

equals  -r,    and    through    D     draw   the    parabolic    curve 


RDFP,  Across  the  figure  draw  a  series  of  ordinates,  as 
CD  and  EF.  Then  any  one  of  these  ordinates  is  equal  to 
d*  or  the  square  of  the  required  depth  of  the  beam  at  the 
location  of  that  ordinate.  To  find  d,  the  depth,  at  each  of 
these  points,  we  have  but  to  make  CG  equal  to  the  square 
root  of  CD,  and  EH  equal  to  the  square  root  of  EF,  and 
in  like  manner  find  corresponding  points  to  G  and  H  on 
each  ordinate,  and  draw  the  curve  line  RGHP  through 
these  points  ;  then  this  curve  line  will  define  the  top  edge  of 
a  beam  (RP  being  the  bottom  edge),  which  shall  be  equally 
strong  at  all  points  to  bear  safely  the  equably  distributed 
load. 


164  STRAINS  FROM  UNIFORMLY  DISTRIBUTED  LOADS.    CHAP.  X. 

I 


T 


P      f 


FIG.  43. 

216.—  The   Form  of  Side  of  Beam  a    Semi-el  lip§e.  —  The 

form  of  the  top  edge  of  the  beam  as  obtained  in  the  last 
article  is  elliptical,  as  may  be  shown  thus  : 

The  equation  to  the  ellipse,  the  co-ordinates  taken  as  in 
Fig.  43,  is* 

12 

U2  =  -  (2(IX  —  X2) 

in  which  x  (=  RE,  Fig,  43)  is  the  abscissa,  u  (=  EH)  is  its 
ordinate,  a(=RC=%f)  is  the  semi-transverse  diameter, 
and  b  (—  CG  —  \/~CJ})  is  the  semi-conjugate  diameter: 
therefore  If  =  CG*  =  CD  and,'  by  formula  (72.\  in  which 
CD,  the  height  of  the  parabola  at  the  middle  in  Figs.  42  and 
43,  is  represented  by  y,  at  its  maximum  we  have  y  =  \Ul. 
In  the  above  value  of  u*  substituting  for  a,  and  b,  their 
values  as  here  shown,  we  have 


and  since    Ix  —  x'  =  x  (I  —  x)  =  th    of  Fig.  42,  therefore 


By  referring  to  formula  (71.)  it  will  be  seen  that  this  value  of 
ua  is  identical  with   that  given  for  y,  the  ordinate  to  the 


*  Cape's  Mathematics,  Vol.  II.,  p.  21,  putting  ;/  for;-. 


FORM    OF   BEAM  AN    ELLIPSE. 


I65 


parabola,  consequently  y  —  u*,  and  therefore  the  curve 
RGHP  is  elliptical. 

To  obtain  the  shape  of  the  beam,  instead  of  drawing  a 
series  of  ordinates  in  a  parabola,  and  taking  the  square  root 
of  each  ordinate,  we  may  at  once  draw  the  semi-ellipse 
RGHP. 

Formula  (73.)  gives  the  value  of  d*  at  middle,  therefore 
for  d  at  middle  make  CG,  Fig.  44,  equal  to 


-\/'~Ual 
"V  ~ 


(U) 


and  through    RGP    draw  a  semi-ellipse,  then    RGPCR    will 
be  the  shape  of  the  beam. 


7? 


FIG.  44. 

As  an  example  : — With  a  beam  of  white  pine  10  feet  long, 
5  inches  broad,  and  loaded  with  10,000  pounds  equably  dis- 
tributed, and  with  a  factor  of  safety  a  =  4,  what  should  be 
the  height  at  the  middle? 

Formula  (74-)  becomes 


10000  x  4  x  10 

2  X  500  X  5 


=  8-94 


or  the  height  of  the  beam  is  to  be  9  inches,  and  the  form  of 
the  side  is  to  be  that  of  a  semi-ellipse,  with  10  feet  for  its 
transverse  diameter,  and  9  inches  for  its  semi-conjugate 
diameter. 


166  STRAINS  FROM  UNIFORMLY  DISTRIBUTED  LOADS.   CHAP.  X. 


QUESTIONS   FOR   PRACTICE. 


2(7. — In  a  scale  of  strains  for  an  equally  distributed  load, 
what  curve  forms  the  upper  edge  ? 

218. — In  a  beam,  10  feet  long,  having  1000  pounds 
equably  distributed  over  its  length,  what  are  the  strains  at 
2,  3,  and  4  feet  respectively,  from  one  end  ? 

219. — What  should  be  the  depth  at  the  middle  of  this 
beam,  if  it  be  of  white  pine,  if  the  breadth  be  made  equal  to 
•fff  of  the  depth,  and  if  4  be  the  value  of  the  factor  of  safety  ? 

220. — In  order  that  the  beam  be  of  equal  strength 
throughout  its  length,  of  what  form  should  the  upper  edge 
be  when  the  lower  edge  is  straight,  and  the  beam  of  parallel 
breadth  throughout  ? 


CHAPTER  XL 


STRAINS   IN   LEVERS,    GRAPHICALLY   EXPRESSED. 


ART.  22 1 « — Scale  of  Strains  for  Promiscuously  Loaded 
I^ever. — In  Fig.  45  we  have  a  semi-beam  loaded  promiscu- 
ously with  the  concentrated  weights  A,  B,  C  and  D. 


FIG.  45. 

To  construct  a  scale  of  strains  for  this  case,  make  EF,  by 
any  convenient  scale,  equal  to  the  product  of  the  weight  A 
into  the  distance  EK\  make  FG  equal  to  BxEU\  make 
GH  equal  to  CxEV',  and  HJ  equal  to  DxET.  From 
each  weight  erect  a  perpendicular,  join  K  and  Fy  L  and  G, 
M  and  //,  and  N  and  J\  then  any  vertical  ordinate,  as 
QP  or  ^5,  drawn  from  the  line  EK  to  the  line  JNMLK, 
will,  when  measured  by  the  same  scale  as  that  with  which 
the  points  F,  G,  H  and  J  were  obtained,  give,  at  the  loca- 
tion of  the  ordinate,  the  effect  produced  by  the  four  weights. 


168     STRAINS  IN  LEVERS,  GRAPHICALLY  EXPRESSED.    CHAP.  XI. 


In  the  construction  of  this  figure,  each  triangle  of  strains 
is  made  upon  the  principle  shown  in  Art.  168,  and  the  several 
triangles  are  successively  added.  An  ordinate  crossing  all 
these  triangles  must  necessarily  be  equal  to  the  sum  of  the 
strains  at  its  location  caused  by  all  the  weights. 

The  strain  at  any  ordinate  may  also  be  found  arithmeti- 
cally, by  taking  the  sum  of  the  products  of  each  weight  into 
its  horizontal  distance  to  the  ordinate,  measured  from  the 
weight  towards  the  wall ;  those  weights  which  occur  between 
the  ordinate  and  the  wall  not  being  considered,  as  they  add 
nothing  to  the  strain  at  the  ordinate. 

222. — Strains  and  Sizes  of  Lever  Uniformly  Loaded. — 

When  the  weights  are  equably  distributed  over  a  semi-beam, 
the  equation  to  the  curve  CFA,  Fig.  46,  limiting  the  ordi- 


II 


FIG.  46. 

nates  of  strains,  may  be  found  by  the  use  of  the  calculus,  as 
in  Art.  212  ;  for  if  ABHJ  be  taken  to  represent  the  equa- 
bly distributed  load,  then  in  considering  the  effect  at  the 
wall  of  a  very  thin  slice  of  this  load,  as  EG  (reducing  it 
infinitely)  we  obtain  the  differential  of  the  strain. 

Let  AE=y,  then  dy,  its  differential,  may  be  taken  as 
the  thickness  of  the  thin  slice  of  the  load  at  EG,  when 
reduced  to  its  smallest  possible  limits.  Putting  e  for  the 
weight  of  a  lineal  foot  of  the  load,  then  edy  will  equal  the 
weight  of  the  thin  slice.  The  effect  or  moment  of  this  slice 


STRAINS   IN   LEVER   COMPUTED   BY   CALCULUS.  169 

at  the  wall,  equals  its  weight  into  its  distance  from  the  wall, 
therefore  we  have  for  the  differential  of  the  moment 

edy  x  (n—y)  •=  du  or, 

endy—eydy  =  du 

The  integral  of  this  expression  is  (Arts.  462  and  4-63) 

/  (endy—eydy)  =  eny—^ey*  =  u 

Applying  this,  or  integrating  between  y  equals  zero  and 
^   equals   n,   we  have 

ert—%evt  =  \en*  —  BC  =  u 
or  for  the  strain  at  the  wall,    BC, 

u  =  \en*  (75.) 

and  for  the  strain  at  any  point,  E, 

x  =  \ey>  (76.) 

From  this  latter,  by  transposing,  we  have 


which  is  the  equation  to  the  parabola  •*  a  proof  that  the 
curve  CFA  is  that  of  a  semi-parabola,  in  which  A  is  the 
apex,  and  CD  the  base. 

These  considerations  pertain  to  the  scale  for  strains.  A 
scale  for  depths  may  be  had  by  proceeding  as  follows  : 

The  value  of  e  in  formulas  (75.)  and  (76.)  is,  from  U  =  en 
(in  which  U  equals  the  whole  load  upon  the  semi-beam) 


*  For,  putting   -  =  /,    then   y2  =  -x  becomes   y2  =  2/je,    the  equation  to 
the  parabola.     See  Cape's  Mathematics,  Vol.  II.,  p.  47. 


STRAINS  IN  LEVERS,  GRAPHICALLY  EXPRESSED.    CHAP.  XI. 

e  =  — .     Substituting  this  value  for    e    in  formula  (75.)  we 
n 

have 

u  =  %--n2 
n 

Putting  this  equal  to  the  resistance  (Art.  35)  gives  us 

=  Sbd* 


and  substituting  for   5  its  equivalent   \B,   and  inserting  the 
symbol  for  safety  (Art.  73),  we  have 

4*4  £/»  =5  Bbd*  or, 

2Uan  =  Bbd2 

[which  agrees  with  formula  (#0.)]  for  the  size  of  the  semi- 
beam  at  the  wall. 

Again,  subjecting  formula  (76.)  to  like  changes,  we  have 
for  the  size  of  the  semi-beam  at  any  point 

2U-y*  =  Bbd2  (77.) 

in  which  y  is  the  distance  of  that  point  from  the  free  end  of 
the  semi-beam. 

223. — The  Form  of  Side  of  L,ever  a  Triangle. — If  a  semi- 
beam,  subjected  to  an  equally  distributed  load,  be  of  rect- 
angular section  throughout,  and  of  constant  breadth,  then,  in 
order  that  it  may  be  equally  strong  at  all  points  of  its  length, 
the  form  of  its  side  must  be  a  triangle. 

This  may  be  shown  as  follows  : 

Formula  (77.)  gives  by  transposition 


in  which   the   coefficient  -^r~ ,   for  the  case  above  cited,  is 


FORM  OF  LEVER  FOR  EQUABLY  DISTRIBUTED  LOAD.       I/I 


composed  of  constant  factors  ;  hence  d*  will  vary  as  y*,  and 
therefore  d  will  be  in  proportion  to  y.  From  this,  formula 
(78.)  is  shown  to  be  the  equation  to  a  straight  line,  and  in 
such  form  that  when  y  equals  zero,  d  also  becomes  zero. 
From  this,  the  side  elevation  of  the  semi-beam  must  be  a  tri- 
angle, with  the  depth  at  the  wall  (for  then  y  becomes  equal 
to  n  )  equal  [from  formula  (78.)  or  (00.)]  to 


d  — 


2Uan 


(79.) 


As  an  example,  let  it  be  required  to  define  the  depth  of  a 
semi-beam  of  white  pine,  10  feet  long  and  5  inches  broad, 
carrying  5000  pounds  equably  distributed  along  its  length, 
and  with  a  factor  of  safety,  a,  equal  to  4. 

Formula  (79.)  becomes 


d  =  4/2  X  5000X4  X   10  y 

500  x  5 


This  is  the  depth  at 
the  wall,  as  at  AC,  Fig. 
47,  in  which  AB  is  the 
length  of  the  semi-beam. 
By  joining  B  and  C  we 
have  ABC  for  the  shape 
of  the  side  of  the  required 
semi-beam. 


FIG.  47. 


224-. — Combination  of  Conditions — The  forms  of  strain 
scales  for  loads  under  various  simple  conditions  having  been 
denned,  we  may  now  consider  those  arising  from  combina- 
tions of  conditions. 

225. — Strains  and   I>imen§ioiis    for    Compound    Load. — 

Take  the  case  of  a  semi-beam  or  lever,  carrying  an  equably 
distributed  load,  and  also  a  concentrated  load  at  the  free  end. 


J/2  STRAINS  IN  LEVERS,  GRAPHICALLY  EXPRESSED.  CHAP.  XI 

Let  the  line    AB,    Fig.  48,  represent  the  length  of   the 


FIG.  48. 

lever,  R  a  weight  suspended  from  its  free  end,  and  DC  the 
face  of  the  wall  into  which  the  lever  is  secured.  In  formula 
(75.)  we  have  the  strain  at  the  wall,  in  which  e  equals  the 

weight  per  lineal  foot  of  the  load,  or   e  —   — .    Substituting 

this  value  in  the  formula,  we  have  21  =  \Un  as  the  strain  at 
the  wall;  therefore  make  AD  =  %l7n,  and  by  the  same  scale 
make  AC—  RxAB  =  Rn.  Join  B  and  C,  and  describe  a 
semi-parabola  from  B  to  D  Avith  the  apex  at  B,  and  the 
base  extended  from  D  parallel  with  AB ;  then  any  vertical 
ordinate  drawn  from  the  curve  DB  to  the  straight  line  CB 
will  measure  the  strain  at  the  point  of  intersection  with  the 
line  AB. 

The  scale  here  given  is  that  for  strains ;  the  scale  for 
depths  will  now  be  shown. 

We  have  seen  in  Art.  223  that  the  form  of  the  side  of  a 
lever  required  by  a  uniformly  distributed  load  is  that  of  a 
triangle,  the  vertical  base  of  which  is  determined  by  formula 
(7#.) ;  and  it  is  shown  at  Art.  178,  that  the  form,  for  a  load 
concentrated  at  the  end  of  a  lever,  is  a  semi-parabola,  with 
its  apex  at  the  free  end  of  the  lever,  and  its  base  vertical  at 
the  fixed  end  and  equal  to 


FORM  OF   LEVER  FOR  COMPOUND   LOAD.  173 

Therefore   let  AB,  Fig.  49,  be  the  length  of  the  lever 


FIG.  49. 

secured  at  A  in  the  wall  DC,  and  having  suspended  from 
its  free  end,  B,  the  weight  P,  and  also  carrying  an  equa- 
bly distributed  load  ABEF.  Make,  by  formula  (79.), 

AD  = 

and  join  B  and  D ;  then  ABD  is  the  scale  for  the  depths 
required  by  the  equally  distributed  load  U.  Make,  as 
above, 


AC=V^ 


Bb 

and  upon  AC  as  a  base  and  AB  for  the  height  describe 
the  semi-parabola  ABC,  which  gives  the  scale  for  depths 
due  to  the  concentrated  load  P. 

Now,  an  ordinate  drawn  at  any  point,  as  G,  vertically 
across  the  combined  scales  of  depths,  as  H  to  J,  measures, 
by  scale,  the  required  depth  for  the  lever  at  the  point  G. 

The  length  of  any  ordinate,  as  HJ,  may  be  determined 
analytically  thus.  The  portion  of  the  ordinate  representing 
the  equably  distributed  load  is,  by  formula  (77.), 


2Ua 
~Bbn 


174     STRAINS  IN  LEVERS,  GRAPHICALLY  EXPRESSED.   CHAP.  XI. 

For  the  remaining  part  of  the  ordinate  we  have  formula 
(36.)  (in  which  x  is  equivalent  to  the  y  of  this  case), 


Adding  these  we  have  for  the  full  length  of  the  ordinate 
HJ,   or  for  the  depth  at  the  point    G, 


d  — 


2Ua 


Bbn 


Bb 


y 


(80.) 


in  which  U  is  the  weight  equably  distributed  over  the 
length  of  the  lever ;  P,  the  weight  concentrated  at  the  end 
of  the  lever ;  #,  the  length  of  the  lever ;  j/,  the  horizontal 
distance  from  the  free  end  of  the  lever  to  the  location  of  the 
ordinate  at  which  the  strain  is  being  measured  ;  a,  the  factor 
of  safety ;  b,  the  breadth  of  the  lever,  and  B  the  resistance 
to  rupture  as  per  Table  XX. 

226.— Scale  of  Strains  for  Compound  Loads. — Fig.  5° 
represents  the  case  of  a  semi-beam  like  the  preceding,  except 
that  the  concentrated  load  is  located  at  some  other  point 
than  the  extreme  end. 


FIG.  50. 

The  curve  DB  is  found  as  in  Fig.  48,  and  the  line  CE 
in  the  same  manner  as  there,  except  that,  in  finding  AC,  the 
distance  m  from  the  wall  to  the  weight  R  is  to  be  substi- 
tuted for  n,  the  length  of  the  lever. 


LEVER   PROMISCUOUSLY    LOADED. 


175 


227« — Scale  of  Strains  for  Pronii§cuoiis  Load. — A  semi- 
beam,  equably  loaded,  may  also  have  to  carry  two  or  more 
concentrated  loads.  In  this  case,  for  the  scale  of  strains  we 
combine  the  methods  required  for  the  two  kinds  of  loads, 
as  in  Fig.  51.  Here  AB  represents  the  length  of  the  semi- 
beam  ;  the  curve  DB,  for  the  equably  distributed  load,  is 


obtained  as  in  Art.  222  ;  and  the  triangles  for  the  concen- 
trated weights  are  found  as  in  Art.  221. 

A  vertical  ordinate  drawn  anywhere  across  the  figure, 
and  terminated  by  the  curve  DB  and  the  line  KJHEB, 
will  measure  the  strain  at  the  location  of  that  ordinate.  The 
depth  of  the  beam  at  that  point  may  be  found  by  putting  the 
strain  as  above  found  equal  to  the  resistance ;  or. 


or  (Art.  35), 
from  which, 


D  =  Sbd* 
D  =  \Bbd'< 


Bb 


in  which  D  represents  the  destructive  energy  or  the  strain 
as  shown  by  the  length  of  the  vertical  ordinate  obtained  as 
above  directed ;  a,  the  symbol  for  safety  (Art.  73) ;  E 
equals  the  resistance  to  rupture  as  per  Table  XX.,  and  b 
and  d  are  the  breadth  and  depth,  respectively — the  breadth 
being  constant. 


STRAINS  IN  LEVERS,  GRAPHICALLY  EXPRESSED.   CHAP.  XL 


QUESTIONS   FOR    PRACTICE. 


228. — In  a  semi-beam  6  feet  long,  carrying  500  pounds 
at  2  feet  from  the  wall,  and  300  pounds  at  5  feet  from  the 
wall,  what  are  the  respective  strains  at  i,  2,  3,  4  and  5 
feet  from  the  free  end  ? 

What  is  the  strain  at  the  wall  ? 

229. — In  a  scale  of  strains  for  a  semi-beam  equably 
loaded,  what  curve  limits  the  upper  edge  ? 

230. — A  semi-beam,  8  feet  long,  is  equably  loaded  with 
100  pounds  per  foot  lineal. 

What  is  the  strain  produced  at    5   feet  from  the  free  end  ? 

231. — Of  Avhat  form  is;  the  side  of  the  last-named  semi- 
beam  required  to  be,  in  order  that  the  beam  may  be  of  equal 
strength  at  all  points,  the  breadth  being  constant  ? 

232. — In  a  semi-beam  7  feet  long,  carrying  1000  pounds 
at  its  free  end,  and  100  pounds  per  foot  lineal,  equably  dis- 
tributed, what  are  the  respective  strains  at  3,  5  and  7  feet 
from  the  free  end  ? 

233. — In  a  semi-beam  10  feet  long,  carrying  an  equably 
distributed  load  of  1000  pounds,  and  concentrated  loads  of 
800,  500  and  700  pounds,  at  the  several  distances  of  3,  6 
and  8  feet  from  the  free  end,  what  are  the  respective  strains 
at  2,  4,  7  and  9  feet  from  the  free  end  ? 


CHAPTER  XII. 

COMPOUND   STRAINS   IN   BEAMS,  GRAPHICALLY   EXPRESSED. 

ART.  234. — Equably  Distributed  and  Concentrated  Loads 
on  a  Beam. — We  have  now  to  consider  the  effect  ot  com- 
pound weights  upon  whole  beams. 

Of  this  class  we  shall  take  first  the  case  of  an  equably 
distributed  weight,  together  with  a  concentrated  one,  as  in 

Fig.  52. 

In  this  figure  the  curve  of  strains  RFDP  for  the  equably 
distributed  load  is  a  parabola,  with  its  apex  at  D.  The 


FIG.  52. 

height    CD  is,  by  formula  (70.),  to  be  made  equal  to  •§-£//; 
and  HJ,  by  the  same  scale,  and  by  Art.  192,  is  to  be  made 

equal    to    A' -j- .    Join   J  with   R  and  with    P.     Then  any 
vertical  ordinate    FG    drawn  across  the  figure,  and  termi- 


1/8    COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

nated  by  the  curve  RFDP  at  top,  and  by  the  line  RJP  at 
bottom,  will  measure  the  strain,  j,  at  E,  the  point  of  inter- 
section of  the  ordinate  with  the  line  RP. 

To  obtain  this  strain  analytically,  we  have,  for  the  ordi- 
nate EF,  formula  (71.),  which  is  (putting  u  for  y  ) 

ujht 
u  =  \U-j- 

and,  for  the  ordinate  EG,  formula  (44*)>  which  is  (putting  b' 
for/,  A'  for  W  and  //  for  x) 


Now,  since  b'+u  —  EG+EF  =  y,    therefore 


^  *,, 

y  =  u+b'  =   \U-j  +  A'-h 

y  =  j(±Ut  +  A'm)  (81.) 

equals  the  strain  at  any  point  between    H  and    P. 

To  find  the  requisite  depth  of  the  beam  at  any  point,  the 
breadth  being  constant,  we  put  the  strain  equal  to  the 
resistance,  or  (Art.  35) 

y  =  Sbd2  =  ±Bbd* 

or,  for  the  safe  weight, 

^ay  =  Bbd*  from  which 

Bbl 

235.  —  Greatest  Strain  Graphically  Represented.  —  To  find 
the  longest  ordinate,  and  consequently  the  greatest  strain, 
arising  from  the  compound  loads  of  Fig.  52,  draw  the  tangent 
KL  parallel  with  JP\  then  an  ordinate  FG  drawn  from 


GREATEST   STRAIN — LOCATION   DETERMINED. 


179 


the  point  of  contact,     F,     will  be  greater  than  any  other 

which  may  be  drawn  across  the  figure. 

• 

236. — Location  of  Greatest  Strain  Analytically  Defined. 

— The  point  of  contact  between  a  curve  and  its  tangent  is  not 
easily  found  by  mere  inspection,  but  analytically  its  exact 
position  may  be  defined. 


FIG.  52. 

To   do  this,  let   (Fig.  52)    a'  =  HJ,    V  =  EG,     u  =  EF, 
h  =  EP    and    h  +  /  =  /  =  RP. 

We  now  have,  from  the  similar  triangles  HJP  and   EGP, 

n  :  a'   : :    h  :  V  =  — 


From  formula  (70.),  in  which   y  =  u  =  \et(l—f),   we  have 
u  =  %eh(l—/i)  =  \ehl-  \eh*  therefore 


— 
n 


(83.) 


This  is  the  value  of  an  ordiriate  drawn  at  any  point  be- 
tween   H    and    P.     But   it  is  required  to  find   where  this 


ISO  COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

ordinate  will  be  at  its  maximum.  This  may  be  done  by 
the  calculus.  Obtain  the  differential  of  formula  (83.\  and 
placing  it  equal  to  zero,  derive  its  integral  ;  from  which  the 
value  of  h  will  be  obtained.  This  represents  the  distance 
from  P  to  the  ordinate  y,  when  at  its  maximum,  and  there- 
fore determines  the  point  E,  the  location  of  the  ordinate,  as 
required. 

237.  —  Location  of  Greatest  Strain  Differentially  Defined. 

—  First.  For  the  value  of  h  we  are  to  find  the  differential  of 
formula  (83.)  and  put  it  equal  to  zero  ;  thus  : 

dy  =  (—  +  \el\dh  —  %e  x  2hdh  =  o 
V;z  / 


=  ehdh 


Now,  since    el=  U,    therefore    e  =  -,-,  and 


Again,  <i=ff?=A'™  therefore 


~  (84.) 


GREATEST   STRAIN   DETERMINED   ANALYTICALLY.         l8l 

or  the  distance  of  the  ordinate  from  the  remote  end  of  the 
beam  is  equal  to  half  the  length  of  the  beam,  plus  a  fraction 
which  has  for  its  numerator  the  product  of  the  concentrated 
weight  into  its  distance  from  the  nearest  bearing,  and  for  its 
denominator  the  weight  which  is  equably  distributed  along 
the  beam. 

This  formula  of  the  value  of  h  is  limited  in  its  applica- 
tion to  those  cases  in  which  n  exceeds  h  in  value.  When, 
on  the  contrary,  k  exceeds  n  ,  then  the  longest  ordinate  is 
at  the  location  of  the  concentrated  weight,  and  n  is  to  be 
substituted  for  h.  The  reason  for  this  may  be  seen  by  an 
inspection  of  the  figure. 

238.  —  Greate§t  Strain  Analytically  Defined.  —  Second.  To 
find  the  length  of  the  ordinate  y,  we  have,  by  formula  (83.), 


n 
and  by  substituting  for    /   its  value,    h  +  t, 


y  - 


a'h 

— 


XT  /  Aimn  j"  U  r 

Now,     a  =  A  -j-  ,  and    e  —  -r,    therefore 


'—  h 


y  — 


n 


1 82   COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

which   gives   the   greatest   strain    resulting   from    both    the 
concentrated  and  distributed  loads. 

This  formula  is  identical  with  formula  (81.),  obtained  by 
another  process. 


239.  —  Example.  —  As  an  example,  let  it  be  required  to 
find  the  location  and  length  of  the  longest  ordinate  of  strains 
produced  by  a  load  of  4000  pounds,  concentrated  at  three 
feet  from  one  end  of  a  beam  16  feet  long,  together  with  a 
load  of  3000  pounds,  equably  distributed  over  its  length. 

First.  The  location  of  the  ordinate,  or  the  value  of  h. 
This,  from  formula  (84-),  is 


or  the  longest  ordinate  is  situated  within  one  foot  of  the 
location  of  the  concentrated  weight. 

Second.  The  amount  of  strain  at  this  ordinate.     This,  by 
the  above  formula,  is 


12 


3+ix  3000x4)  =  13500 


or   the   greatest   resulting   strain    at  any    one   point   of  the 
combined  weights  equals    13,500   pounds. 

240. — IMmeiiiions  of  Beam  for  Distributed  and  Concen- 
trated Loads. — The  amount  of  strain,  just  found  is  the  actual 
moment  of  the  loads.  Putting  this  equal  to  the  resistance 
(Art.  35),  we  have,  for  the  safe  weight, 

a^(A  'm  +££#)  =  Sbd2  =  \Bbd2  or 

/ 

4*  7  (A  'm  +  ±Ut)  =  Bbd*  (85.) 


DIMENSIONS   OF  BEAM   FOR  COMPOUND   LOAD.  183 

which  is  a  rule  for  obtaining  the  dimensions  requisite  for  re- 
sisting effectually  the  greatest  strain  arising  from  the  com- 
bined action  of  a  concentrated  and  an  equably  distributed  load  ; 
and  in  which  A'  equals  the  concentrated  load,  and  U  the 
equably  distributed  load,  both  in  pounds  ;  /  is  the  length  of 
the  beam  between  bearings  ;  m  the  distance  from  the  con- 
centrated weight  to  the  nearer  end  of  the  beam  ;  h  the  dis- 
tance from  the  location  of  the  greatest  strain  to  the  more 
distant  end  of  the  beam  ;  and  /  equals  /  —  //.  /,  m,  h  and 
/  are  all  to  be  taken  in  feet,  and  the  value  of  h  is  to  be  had 
from  formula  (84-} ;  care  being  exercised  that  when  h  ex- 
ceeds n  in  value,  then  n  is  to  be  used  in  place  of  h,  and 
m  in  place  of  /.  In  the  latter  case  formula  (85.)  becomes 

4a  ^(A'm  +  tUm)  =  Bbd2 
=  Bbd* 


24-1. — Comparison  of  Formulas,  Here  and  in.  Art.  ISO. — 
Formula  (29. \  given  in  Art.  150,  for  a  carriage  beam  with  one 
header,  is  for  a  case  similar  to  that  of  the  last  article,  but  is  not 
strictly  accurate.  Instead  of  the  two  strains  being  taken  at 
the  same  point,  E  (the  location  of  the  longest  ordinate),  as 
in  Fig.  52,  they  are  taken,  the  one  for  the  concentrated  load, 
at  the  location  of  this  load,  and  the  other,  that  for  the 
equably  distributed  load,  at  the  middle  of  the  beam ;  or,  the 
maximum  strain  for  each  load. 

Taken  in  this  manner  the  result  is  in  excess  of  the  truth, 
as  //7+  CD  is  greater  than  FG.  The  error  is  upon  the 
safe  side,  the  strains  being  estimated  greater  than  they  really 
are.  In  most  cases  this  error  would  not  be  large,  and  the  only 
objection  to  it  would  be  that  it  requires  a  little  more  mate- 
rial in  the  beam.  Formula  (29.)  may  therefore  be  employed 


1 84  COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

in  ordinary  cases  where  a  low  priced  material,  such  as  wood 
for  example,  is  used  for  the  beams ;  but  where  a  more  costly 
material  is  involved,  economy  would  dictate  that  the  strain 
be  not  over-estimated,  and  that  it  be  correctly  obtained  by 
the  use  of  formula  (85.)  in  Art.  24-0.  (See  also  caution  in 
Art.  88.) 

242.— Location  of  Oreate§t  Strain  Differentially  De- 
fined.— In  Fig.  53  we  have  a  scale  of  strains,  RABPF,  by 
which  is  found  the  effect  arising  at  any  point  in  the  length  of 
the  beam  from  two  concentrated  loads,  together  with  an 
equably  distributed  load. 

The  curve  RFP  is  a  parabola  (foot  note,  Art.  2(2) 
found  as  in  Fig.  52,  and  the  moment  of  the  two  concen- 
trated loads  equals  AH  at  H  and  BJ  at  J,  and  is 
found  as  in  Art.  194-  and  Fig.  37.  FG  is  the  ordinate  for 
strains  occurring  between  H  and  J1  and  is  defined  thus  : 

.L 


Let      HJ  =  d'         EJ  ±±  x,       EG  =  v  =  b'+fr       EF  =  «, 


AH=af,     BJ=V     and    a'—bf  =  c.      Then,    from     similar 
triangles, 


GREATEST  STRAIN — LOCATION  BY  CALCULUS.     185 


and,     since     x  =  h— s,      v  =  b'+p    and      c  —  a'— b't     there 
fore 


d' 

af-V 
and  v  —  b  -\ 77—  < 


Formula  (70.), 


gives  [putting  ///  for  t(l—t)    and   u   for  y\ 

u  =  %eht 
and  since  y  —  u  +  v,  consequently 

y  =.  ^eht  +  b'  -\ -j, —  (h — s)  (87.) 


This  is  the  value  of  the  ordinate  for  the  strain  at  any 
point  between  H  and  J. 

To  obtain  the  longest  ordinate  which  can  be  drawn  here, 
proceed  as  in  Arts.  235  to  237,  and  as  follows: 

First  reduce  formula  (87.)  thus, 


a'-b'          .       a'-b'         a'-V 


then         v  +  u  =  y  =  \ehl—^ehs  +  b'  -i  —  -r,  —  h  --  -«  —  s 

In  this  expression,  rejecting  the  quantities  unaffected  by  the 
variable   h,    we  have,  for  the  differential  of  y, 


1  36  COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.   CHAP.  XII. 
dy  =  (^el  +  ^^dh  -  ehdh  =  o 


or,  (  \el  +  -—«—  }dk=.  ehdh 

or,  its  integral  gives 

*  =  #+£=  (**.) 


243.—  Greatest  Strain  and  Dimensions  —  The  above  gives 
the  value  of  h.  To  obtain  the  value  of  y  at  its  maximum 
take  formula  (87.).  In  this,  for  the  value  of  a'  we  have  AH, 
equal  to  the  joint  effect  at  H  of  the  two  concentrated  loads  ; 
or,  putting  a'  for  the  D  of  formula 


m 


and  for  the  value  of   b'    (form.  52.) 

b'  =  j(B'r  +  A'm) 

The  value  of   e  (from   el—  U)  is  equal  to   7-  .      By  sub- 
stituting this  value  for  e  we  have 

=U      +  y+k-s  (89.) 


This  equals  the  strain  from  the  compound  weights  of  Fig.  S3, 
and  is  the  same  as  (87.),  for    —j  ==  \e. 

Either  formula  will  give  the  strain  at  any  required  point 
between    H   and    J   (Fig.  53)  by  putting    h    equal  to  the  dis- 


DIMENSIONS   OF  BEAM    FOR  COMPOUND   LOAD.  1 87 


FIG.  53. 

tance  between  that  point  and  P  ;  but  when  the  greatest 
strain  is  required,  h  must  be  obtained  from  its  value  in 
formula  (88.).  To  obtain  the  dimensions  in  this  case,  we  put 
the  strain  equal  to  the  resistance,  and  have,  with  a  as  the 
factor  for  safe  weight  (Arts.  35  and  73) 


~        =  Sbd*  = 


=  Bbd* 


and  from  this  formula  may  be  found  the  dimensions  required 
for  resisting  effectually  the  greatest  strain  in  the  beam,  the 
value  of  h  being  derived  from  formula  (88.). 

24-4.— A§§ignlng  the  Symbol*. — It  is  important  to  observe 
here  that  of  the  two  moments  a'  and  £',  a'  designates  the 
larger  of  the  two,  while  m  and  n  represent  the  distances 
from  a'  to  the  two  ends  of  the  beam,  m  being  the  distance 
to  that  support  which  may  be  reached  without  passing  the 


1  88   COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.   XII. 

other  weight.  Again  r  and  s  are  to  be  regarded  as  the 
distances  from  b'  to  the  two  ends  of  the  beam  ;  k  and  s 
dating  from  the  same  end  of  the  beam  as  n  ;  and  as  n  is  the 
greatest  possible  value  of  //,  it  is  to  be  substituted  for  it 
when  by  the  formula  for  h  its  value  is  found  equal  to  or 
greater  than  n. 

In  order  to  ascertain  which  of  the  two  moments  a  and 
b'  is  the  greater,  a  trial  must  be  had  by  the  use  of  the  expres- 
sions in  the  last  article  designating  their  respective  values. 
When  the  two  concentrated  weights  are  equal,  then  the 
nearer  weight  to  the  middle  of  the  beam  will  produce  the 
greater  moment,  and  may  at  once  be  designated  as  a. 

245.—  Example—  Strain   and  Size  at  a  Given  Point.  —  As 

an  example,  let  a  beam,  10  feet  long,  be  required  to  carry  an 
equably  distributed  load  of  100  pounds  per  foot  lineal,  a  con- 
centrated load  of  2000  pounds  at  a  point  two  feet  from  the 
left-hand  end,  and  a  second  concentrated  load  of  800  pounds 
located  at  3  feet  from  the  right-hand  end.  What  will  be  the 
resulting  strain  at  4  feet  from  the  right-hand  end  ? 
Formula  (87.)  is 


equals  the  required  strain. 

In  designating     m    and     s    we  find    (Art.  243)  for  the 
larger  weight 


—  (2000  x  8  +  800  x  3)  =  3680 


and  for  the  smaller 


—  (8OO  X  7  +  2000  X  2)  =  288O 


and  hence  (Art.  153)     m  =  2     and     s  =  3. 


DIMENSIONS   AT   A    GIVEN   POINT.  189 

We    now    have     e  =  100,    h  =  4,     /=  10,     t  =  l—h  =  6, 
—  2,     n  =  S,     J=3,     r  =  7   and     ^'=5. 

becomes   £  x  100  X4  x  6  =1200 

With     ^'  =  2000     and     B'  =  800,      a'  =  ™(A'n+B's)  = 

(as  above)    3680,    and     bf  =  j  (Br  r+A'  m)  =  (as  above)    2880 

and     #'—  £'  =  3680—2880  =  800. 

We  therefore  have,  as  a  resulting1  value  of    y    in  formula 


1200  +  2880  +  -—(4—3)  =  4240  —  y 

This  equals  the  effect  at    4    feet  from  the  right-hand  end  pro- 
duced by  the  three  weights. 

To  find  the  dimensions  of  the  beam  at  this  point,  make 
the  strain  just  found  equal  to  the  resistance  [see  Art.  24-3  at 
formula  (90.)],  and  we  have 

4a  x  4240  =  Bbd* 
and,  if    a  =  4    and     B  =  500     (see  Table  XX.),  we  have 


500 

Let  £=3,  then  we  have  d  =  6-j$\  or,  the  beam  at 
4  feet  from  the  right-hand  end  should  be  3  x  6-  73  inches  in 
cross-section. 

246.—  Example—  Greatest  Strain.  —  Again,  let  it  be  re- 
quired to  show  the  greatest  strain  produced  at  any  one  point 
by  the  three  weights  of  the  last  article. 

The  first  dimension  required  here  is  that  of  h.  For 
this  we  have,  as  per  formula  (88.\ 


190   COMPOUND  STRAINS,   GRAPHICALLY  EXPRESSED.    CHAP.  XII. 


from  which  h=  $  -^  ---    -  =6-6 

This  result  being  less  than    n,   since    n   equals    8,   is  there- 
fore the  correct  value  of    h,     and  from  it  we  obtain  (from 
—  /)    /—  3.4.     Formula  (89.}  now  gives 


'= 
which  is  the  required  greatest  strain. 


247.—  Example—  Dimensions.  —  What  sized  beam  of  equal 
cross-section  throughout  would  be  required  to  carry  safely 
the  loads  upon  the  beam  of  the  last  article,  when  B  =  500 
and  a  =  4? 

The  greatest  strain  at  any  point  was  found  to  be  4578 
pounds,  therefore 

4a  x  4578  =  Bbd9 
4.X4X4578 

500 

and  with     b    taken  equal  to   3,  then     d=6>gQ.     The  beam 
must  be    3x7   inches. 

248.—  Dimensions  for  Greatest  Strain  when  //  Equals  n.  — 

When,  in  formula  (90.),     h  =  n,     or  is  greater  than   nt    then 
t  =  m,    h—s  =  df  ,  and 


c*f      =  b'+a'-b'  =  a' 
also, 


DIMENSIONS   OF   BEAM  —  COMPOUND   LOAD. 

and  the  formula  becomes 


or,  supplying  the  value  of  a'   (Art.  243), 


which  is  a  rule  for  a  beam  carrying  two  concentrated  loads 
and  a  uniformly  distributed  load,  when  h  =  n  as  above 
stated. 


249.— Dimensions  for  Greatest  Strain  when  h  is  Greater 
than  n. — As  an  example  under  this  rule,  what  are  the 
breadth  and  depth  of  a  Georgia  pine  beam  20  feet  long, 
carrying  2000  pounds  uniformly  distributed  over  its  whole 
length,  10,000  pounds  at  7  feet  from  the  left-hand  end,  and 
8000  pounds  at  5  feet  from  the  right-hand  end  ;  the  factor 
of  safety  being  4  ? 

Here  a  =  4,  [7=  2000,  /=2O,  ^=850,  m  =  7  and 
s  =  5  (since  7  x  10,000  =  70,000  exceeds  5  x  8000  =  40,000 ), 
#=13,  r=i5  and  d'  =  8.  The  value  of  //  is  to  be 
tested,  to  know  whether  it  is  equal  to  or  greater  than  n. 

By  formula  (88.),  and  Art.  243, 

a'—b'  a'-br 

dr 


a'  =  -j-(A'n+B's)  —  —(10000x13  +  8000x5)  =  59500 


V  =  j(B'r+A'iri)  —  —  (Sooox  1  5  +  10000x7)  =  47500 


192    COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

a'  —  b'  =  59500  —  47500=  12000 

12000 


20 

This  gives  a  value  to  h  greater  than  that  of  n  and  shows 
(Art.  24-4)  that  n  must  be  substituted  for  //,  and  that 
the  problem  is  a  proper  one  for  solving  by  formula  (91.)\ 
therefore 

r—  7X13  7     __  _  ~] 

4x4   looo-  —  --  +  —-(10000x13+8000x5) 
'      -  ~~  ="  =  1205.  65 


If  the  breadth    b    be  taken  at    8    inches,  then   ^= 
that  is,  the  beam  should  be     8  x  i2|     inches. 


250.—  Rule  for  Carriage  Beams  with  Two  Headers  and 
Two  Sets  of  Tail  Beams.  —  By  proper  modifications,  formula 
(90.)  may  be  adapted  to  the  requirements  of  a  carriage  beam 
with  two  headers,  as  in  Fig.  25.  These  modifications  are  as 
follows  :  By  Art.  150  we  have 


hence  U~,  =  \cfht 


also,  from  Arts.  153  and  243, 


a   = 


and,  from  Art.  (55, 

A'  =  \fgm    and    B'  =  \fgs 
therefore 


CARRIAGE   BEAM   WITH   TWO   HEADERS.  193 

m 


Similarly  we  find 

b' 


To  obtain  the  maximum  strain,     h    is  to  be  determined 
by  formula  (88.\  in  which  for    e     we  have 

U      cfl 

<-----[----Ti= 

and  therefore 

a'-V 


In  these  deductions,  f  equals  the  weight  per  superficial 
foot  of  the  floor,  c  the  distance  apart  from  centres  at 
which  the  beams  in  the  floor  are  placed,  and  g  the  length 
of  the  header.  (For  cautions  in  distinguishing  between  m 
and  s,  and  between  a'  and  b't  see  Art.  244.)  By 
formula  (90.)  and  the  modifications  proposed,  we  therefore 
have 

Bbd* 


and  as  auxiliary  thereto  we  have,  as  above, 


a'  =~ 


and 

a'-b' 


IQ4  COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

and  thus  we  have  in  formula  (92.)  a  rule  for  a  carriage  beam 
carrying  two  headers  and  two  sets  of  tail  beams.  (See  cau- 
tion in  Art.  88). 


251. — Example. — To  show  the  application  of  this  rule, 
let  it  be  required  to  find  the  breadth  of  a  white  pine  carriage 
beam,  20  feet  long  and  10  inches  deep;  the  beam  to  carry 
two  headers  10  feet  long,  one  located  9  feet  from  the  left- 
hand  end,  and  the  other  6  feet  from  the  right-hand  end.  The 
floor  beams  are  to  be  placed  15  inches  from  centres,  and  the 
floor  is  to  carry  100  pounds  per  superficial  foot,  with  the  fac- 
tor of  safety  a  —  4. 

Here  the  header  at  the  left-hand  end  is  the  nearer  of  the 
two  to  the  middle  of  the  carriage  beam,  and  therefore  (Art. 
244)  m  —  9. 

From  formula  (92.)  we  have,  for  the  value  of  b, 


in  which  a  =  4,    B  —  500,    d*  —  io2,  /=  100,  c  =  ij,  g—  10, 
I—  20,    m  =  q,    n—  \\,    r  =  14,    s  =  6    and    d'  =  5. 
From  the  auxiliary  formulas- of  Art.  250, 


a'  =  100  x  IQX  -— — (9  x  ii  +  62)  =  15187-5 
4  x  20 


b'  =  ico  x  io  x  ^— 2-  (14  x  6  +  92)  ==  12375 

a'—b1  =  15187-5  —  12375  =  2812-5 

2812-5    _ 
/'-=IO  +  "" 


RULE   FOR   CARRIAGE   BEAMS. 


Here  «,  since  it  is  but  n,  is  less  in  value  than  h,  and 
must  be  used  in  its  place  ;  we  therefore  have  recourse  to 
formula  (91-),  Art.  248.  By  this  formula  the  value  of  b  is 


This  is  a  general  rule.  To  make  it  conform  to  the  re- 
quirements for  a  carriage  beam,  we  have  for  U  the  equally 
distributed  load  \cfl  (Art.\BO). 

A'  =  \fgrn   (Art.  250),  and   B'  =  Ifgs.     Hence 


=  I^r  +  g 

SooxiooL  20  ^  JJ 

or,  the  carriage  beam  should  be  5^  or,  say  6  inches  broad. 
In  this  computation,  no  allowance  is  made  for  the  weaken- 
ing effect  of  mortising,  it  being  understood  that  no  mortises 
are  to  be  made  ;  the  headers  being  hung  in  bridle  irons 
(Art.  147).  (See  Art.  88). 


252. — Carriage  Beam  with  Three  Header*. — It  some- 
times occurs  in  the  plan  of  a  floor  that  two  openings,  the  one 
a  stairway  at  the  wall,  the  other  an  opening  for  light  at  or 
near  the  middle  of  the  floor,  are  opposite  each  other,  as 
in  Fig.  54. 

In  this  arrangement  the  carnage  beam  has  three  headers 
to  carry,  besides  its  load  as  an  ordinary  floor  beam. 


196  COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 


FIG.  54. 

Cases  of  this  kind  may  be  divided  into  two  classes :  one 
that  in  which  the  header  causing-  the  greatest  strain  occurs 
between  the  other  two  ;  the  other,  that  in  which  it  occurs  next 
to  one  of  the  walls.  We  will  first  consider  the  latter  case. 


253. — Three  Headers— Strains  of  the  Fir§t  Class. — When 
the  well  hole  for  light  occurs  at  the  middle  of  the  distance 
between  the  walls,  its  two  headers  will  be  equally  near  the 
centre  of  the  length  of  the  carriage  beam  ;  and,  were  their 
loads  alike,  the  headers  would  produce  equal  strains  upon  the 
carriage  beam  ;  but  the  loads  are  not  alike,  for  the  tail  beams 
carried  by  one  header,  those  which  reach  to  the  wall,  are 
longer  than  those  carried  by  the  other. 

Hence  the  header  carrying  the  tail  beams,  one  end  of 
which  rest  on  the  wall,  has  the  heavier  load  ;  and,  as  it  has 
the  same  leverage  as  the  header  on  the  other  side  of  the 
well  hole,  it  will  therefore  have  the  greater  moment,  and 
will  produce  the  greater  strain  upon  the  carriage  beam. 


CARRIAGE  BEAM  WITH  THREE  HEADERS. 


I97 


The  stair  header  will  add  to  the  strains  upon  the  carriage 
beam  at  the  points  of  location  of  the  other  two  headers,  and 
this  addition  will  be  greater  at  the  middle  header  than  at  the 
farther  one,  but  still  not  so  much  greater  as  to  cause  the 
total  strains  at  the  one  to  preponderate  over  those  at  the 
other. 


254.—  CJrapIiical  Representation. — Let  Fig.  55,  construct- 
ed similarly  with  Fig.  53,  represent  the  strains  in  a  carriage 
beam  supporting  three  headers,  one  of  the  outside  ones,  as  at 
A,  producing  the  greatest  strain.  In  this  figure  the  curve 
DKE  is  a  parabola  (Art.  212)  and  is  the  curve  of  strains 
for  the  uniformly  distributed  load  upon  the  carriage  beam, 


FIG.  55. 

of  which  KL  represents  the  strain  at  the  middle  of  the 
beam ;  and  CF,  BG  and  AH,  vertical  lines,  by  the 
s'ame  scale,  represent  the  strains  caused  by  the  three  head- 
ers at  the  points  C,  B  and  A,  respectively.  Any  or- 
dinate  drawn,  parallel  to  AH,  across  this  figure,  and  ter- 
minated by  the  boundary  line  DFGHEKD,  will  measure 
the  strain  in  the  carriage  beam  at  its  location.  Hence 
that  point  at  which  an  ordinate  thus  drawn  proves  to 
be  longest  of  any  which  may  be  drawn,  is  the  point  where 
the  strain  upon  the  carriage  beam  is  the  greatest,  and  the 
length  of  this  ordinate  measures  the  amount  of  this  strain. 


IQ8    COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 


Draw  the  tangent 


ST  parallel  to  GH.  If  its  point  of 
contact  with  the  curve  occurs  between  Q  and  £,  then 
HQ  will  be  the  longest  possible  ordinate  ;  but,  if  it  occur 
between  K  and  Q,  then  HQ  will  not  be  the  longest. 
When  AH  and  BG  are  equal,  the  point  of  contact  will 
be  at  K.  In  the  case  under  consideration  (the  well  hole  in 
the  middle  of  the  floor)  the  tangent  will  usually  touch  be- 
tween Q  and  £,  giving  HQ  as  the  longest  ordinate. 


255.—  Greateit  Strain.  —  With  the  loads     A,    B     and     C 
in  position  as  in  Fig.  55,  the  longest  ordinate  may  be  found 


FIG.  55. 
by  formula  (87.),  where 

/ ri 

y  =  \eht  +  b'  -i — -77 —  (h—s) 


and  in  which     m  +  n  =  r+s=/i  +  t  =  t    (for  the  position  of 
these  letters  see  Art.  244),  \eht    represents  the  strain  from 


distributed      load,     and 


a'~^-(h     s) 
//       v» */ 


the    uniformly 

stands  for  the  length  of  an  ordinate  drawn  from  GH  to 
BA  at  the  distance  h  from  D  towards  A,  and  repre- 
sents at  the  location  of  the  ordinate  the  strain  from  the  three 
concentrated  loads.  In  all  cases,  except  where  b'  is  very 
nearly  or  quite  equal  to  a',  h  will  exceed  »,  and,  in 


GENERAL   RULE   FOR  COMPOUND   LOAD.  199 

general,  for  all  problems  of  the  class  of  which  we  are  treat- 
ing, it  may  be  assumed,  without  material  error,  that  h  will 
always  exceed  n.  Then  m  and  n  take  the  place  of  t 
and  h  in  formula  (87.),  and  it  becomes  (Art.  248) 


y  =  \ernn  +  a' 
mn 


or, 


The  value  of    a'     is  (form.  58.) 


a'  =  ~ 


hence  y  =  %  u^  +  j(A  'n  +  B's  +  Cv)  (95.) 

In  this  formula    y    equals  the  greatest  strain  in  the  beam. 

256.—  General  Rule  for  Equably  E>i§tributecl  and  Three 
Concentrated  Loads.  —  Putting  the  strain  y  of  last  article 
equal  to  the  resistance  (Art.  35)  gives  us 

%  Uni~  +  ™(A  'n  +  B's  +  Cv)  =  Sbd* 
and  with     B  —  ^S     and     a     as  the  coefficient  of  safety, 

'n  +  B's  +  C'v)  =  Bbd2         (96.) 


which  is  a  general  rule  for  beams  carrying  a  uniformly  dis- 
tributed load  and  three  concentrated  loads  similarly  placed 
with  those  in  Fig.  55.  In  this  rule,  U  is  the  uniformly  dis- 
tributed load,  and  Af,  B'  and  C  the  three  loads  concen- 
trated at  A,  B  and  C  in  the  figure. 

257.—  Example.  —  As  an   example,  we  will  ascertain   the 
required  breadth  of  a  Georgia  pine  beam  of  average  quality, 


200   COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

20  feet  long  and  14  inches  deep,  with  a  load  of  2000  pounds 
equally  distributed  over  its  length,  a  concentrated  load  of 
4000  pounds  at  3  feet  from  the  left-hand  end,  a  like  load  at 
7  feet  from  the  same  end,  and  one  of  7000  pounds  at  7  feet 
from  the  right-hand  end.  Take  as  the  factor  of  safety  a  =  4. 
Then  /=  20,  ;;/  =  7,  ;/  =  13,  s  =  7,  r  —  13,  v  —  3,  u  =  17, 
d—  14,  U  —  2000,  A'  —  7000,  ^  =  4000  =  C'  and  B  —  850, 
and  from  formula  (96.) 


/  ----      --     -       ---  \ 

b  =  85oxTo(*X200°  X  J3  +  7oooxi  3  +  4000x7  +4000x3^=4.84 

or  the  breadth  should  be  4^  inches. 

258.—  Rule  for  Carriage  Beam§  with  Three  Heatler§  and 
Two  Sets  of  Tail  Beams.  —  To  modify  formula  (96.)  so  as  to 
make  it  applicable  to  a  carriage  beam,  we  have  for  £7,  the 
uniformly  distributed  load,  (Art.  150)  U=%cfl;  for  the 
load  at  A,  caused  by  the  header  carrying  the  tail  beams, 
one  end  of  which  rests  upon  the  wall,  A  '  —  \fgm  ;  for  the 
load  at  B,  B'  —  \fg(s—v)  ;  and  for  the  load  at  C  the 
same,  C'  =  :kfg(s—v).  Formula  (96.)  now  becomes 


b  = 

b  =  TM  ^cnl  +  gmn  +  g  ^~ 

b  =  -^  [cnl+g  (mn  +  s*-v*)]  (97.) 

which  is  a  rule  for  carriage  beams  carrying  three  headers  and 
two  sets  of  tail  beams,  located,  as  in  Fig.  55,  with  A,  the 
heaviest  strained  header  in  an  outside  position  relative  to 
the  other  two  headers. 

259.—  Example.  —  Under  the  above  rule,  what  should  be 
the  breadth  of  a  spruce  carriage  beam    20    feet  long  and    12 


CARRIAGE  BEAM  WITH  THREE  HEADERS. 


2O I 


inches  deep,  carrying  three  headers  15  feet  long,  located  as 
in  Fig.  54.  The  well-hole  for  light,  in  the  middle  of  the  width 
of  the  floor,  is  6  feet  wide,  and  the  stairway  opening,  at  one 
of  the  walls,  3  feet  wide.  The  beams  of  the  floor  are  placed 
15  inches  from  centres,  and  are  to  carry  90  pounds  per 
superficial  foot,  with  4  as  the  factor  of  safety. 

1—6 
Here   /  —  20,    m  =  s  =  —  —    -  7,    n  —  13,   v—^g—\^, 

d—\2,   <:=ii,  /=90,    a  —  4     and     ^= 
By  formula  (97.) 


b  = 


or  the  breadth  should  be  3f  inches. 


-32)]  =  3-64 


260.— Three    Headers— Strains    of    the    Second   Class. — 

We  will  now  consider  the  other  class  named  in  Art.  252, 
that  in  which  the  header  causing  the  greatest  strain  occurs 
between  the  other  two. 

The  conditions  of  this  class  of  cases  are  represented  in 
Fig.  56,  in  which  AH=a',  BG  =  b'  and  CF=cf,  repre- 
senting by  scale  the  combined  concentrated  strains  at  A,  B 


and     C    respectively,  and     KL     is  the  strain  at  the  middle 
due  to  the  uniformly  distributed  load.     The  parabolic  curve 


202   COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.   CHAP.  XII. 

(Art.  212)     EKD     and  the  line     DGHFE     form  the  boun- 
daries of  the  scale  of  strains,  as  in  Art.  254. 

For  the  proper  assignment  of  the  symbols  ///,  n,  r,  s,  etc. 
see  Art.  244,  taking  the  two  larger  of  the  strains  of  Fig.  56 
for  the  two  given  in  that  article. 


The  longest  vertical  ordinate  across  the  scale  of  strains 
will  ordinarily  be  at  QH]  the  exceptions  being  when  the 
strain  at  B  is  nearly  or  quite  equal  to  that  at  A.  In  the 
latter  case,  however,  the  diminution  at  QH  will  be  so  small 
that  that  ordinate  may  be  assumed,  without  material  error, 
to  be  the  greatest.  Taking  it  as  the  greatest,  formula  (87.) 
becomes,  as  in  Art.  255, 


261.— Greatest  Strain. — The  manner  of  obtaining  the  value 
of  a',  the  strain  produced  by  the  three  concentrated  loads, 
will  now  be  shown. 

The  strain  at    A,     produced  by  the  load     A',     is  (Art. 


.mn 


56)    A'-j-.     The  strain  at 


*?• 


B,    produced  by    B' ',    is 
of  the  strain  at    B    may  be  had   by  the 


The  effect  at  A 
proportions  shown  in  the  triangles  BGE  and  ARE ;  for 
the  effect  is  proportional  to  the  horizontal  distance  from  E 
(see  Art.  192);  therefore, 


THREE   LOADS  —  GREATEST   STRAIN.  203 


equals  the  effect  of  the  weight    B'     at  the  point    A. 

Also,   for    the   effect  at    A     of  the  weight  at     Cy    the 

effect  of    C'    at    C    being   C  y-,     we  have 

Cuv        C'nuv       C'nv 
u  :  n  :  :  —  -,  —  :    —,  ---  =  —  •=- 
I  lu  I 

equals  the  effect  at    A     of  the  weight  at  C. 

The  joint  effect  at    A     of  the  three  weights  is  therefore 


or,  af  =  ™Afn 


Adding  this  to  the  effect  of  the   uniformly  distributed  load, 


mn 

~,  gives 


"  (P*.) 


This  represents  the  greatest  strain  arising  from  the  uni- 
formly distributed  load  and  the  three  weights  disposed  as  in 
Fig.  56]  A'  at  the  middle  being  the  greatest  strain  and  B' 
the  next  greatest. 

262.— General  Rule  for  Equally  Distributed  and  Three 
Concentrated  Load§. — Putting  the  strain  [form.  (##•)]  in 
equilibrium  with  the  resistance  (Art.  35)  we  have 

***  tii) 

\'n  +  B's)  +  C'^-  =  Sbd'  = 


204  COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

and  with  the  symbol  for  safety  added, 

b  =          Wi  Un  +  A'n  +  B's)  +  C.  'wv\        (99.) 


which  is  a  rule  for  beams  loaded  with  an  equally  distributed 
load  and  with  three  loads  relatively  disposed  as  in  Fig.  56  ; 
A'  being  the  greatest  strain,  and  B'  the  next  greatest,  and 
A'  being  at  the  middle. 

263.—  Example.  —  As  an  example  under  this  rule  :  What 
should  be  the  breadth  of  a  Georgia  pine  beam  of  average 
quality,  20  feet  long  and  12  inches  deep,  carrying  4000 
pounds  uniformly  distributed,  6000  pounds  at  4  feet  from 
the  left-hand  end,  6000  pounds  at  9  feet  from  the  same 
end,  and  7000  pounds  at  6  feet  from  the  right  hand  end  ; 
with  the  factor  of  safety  a  =  4? 

Assigning  the  symbols  to  the  loads  and  spaces  as  in  Fig. 
56,  we  have 

#  =  4,     ^  =  850,    d—  12,    /=  20,     m  =  9,      n—ii,     r  —  14, 
s  =  6,       v  —  4,       U=  4000,       A'  —  6000,      B'  —  7000       and 
C'  —  6000. 
Substituting  these  values  in  formula  (99.)  gives 

b  —  Tt  --  3  ---  [0(^x4000x11  +6000x1  1  +  7000x6)  + 
Ly 


t 

850x12 

(6oooxi  1x4)]  =9-  37 
or  the  breadth  should  be    9f    inches. 

264-.—  A§§igning  the  Symbol*.  —  In  working  a  problem  of 
the  kind  just  given,  it  is  of  prime  importance  to  have  the 
symbols  denoting  the  weights  and  distances  properly  located. 
In  doing  this,  the  first  point  to  settle  is  as  to  which  of  the  two 
classes  (Fig.  55  or  56)  the  case  in  hand  belongs. 

Make  a  sketch,  such  as  Fig.  55  or  56,  according  to  the 
probable  position  of  the  largest  strain,  letter  the  weights  and 


COMPOUND    LOAD — DISTINGUISHING   THE    RULE.  205 

distances  as  there  shown,  and  then  compute  the  three  strains 
by  the  following  formulas. 

For  Fig  55  the  strains  will  be  as  follows  (Art.  195)  : 


At    A,    the  strain     a  = -r(A'n  + B's+ Cv) 
"     B,  "  b'  =  j(A'm  +  B'f)  +  C'^          (101) 

"     C,  "  c'  =  j(A  'm  -f  B'r  +  Cu)  (102.) 

In  the  diagram,  AH  is  to  be  made,  by  any  convenient 
scale,  equal  to  a',  BG  to  b',  and  CF  to  c ,  as  found 
by  these  three  formulas,  and  KL,  the  height  of  the  para- 
bola, is,  by  the  same  scale,  to  be  made  equal  to  %U '  —j—  .  U 

is  the  load  equably  diffused  over  the  beam  ;    A',  B!    and     C' 
are  the  loads  concentrated  at    At   B    and     C    respectively, 
and    /    is  the  span,  or  length  of  the  beam  between  bearings. 
For  Fig.  56  the  strains  will  be  as  follows  : 

At    A,    the  strain     a'  =  ~ (A  'n  +  B's)  +  C'Hj      (103) 

"     B,          "  b'  =  j(Arm+B'r+Cv)         (104) 

v 
n     Q  «  c>  __  __  /^  'ft  i  j^i s 


In  the  case  of  a  carriage  beam  the  loads  A',  B'  and  C' 
in  the  formulas  (100.)  to  (105.)  are  those  from  the  headers  ; 
and  equal  %fgm,  etc.  In  this,  /  and  g  are  constant,  as 
to  the  three  loads  in  any  given  case,  and  m  represents  the 
length  of  one  set  of  tail  beams  ;  consequently  the  loads  A' 
B'  and  C  will  vary  as  the  length  of  the  tail  beams. 

Hence,  in  the  preliminary  work  required  to  ascertain  to 
which  of  the  two  classes  any  given  case  belongs,  it  will  suf 
fice  to  use  simply  the  length  of  the  tail  beams,  instead  of  the 
full  weights  A',  B'  and  C. 


205   COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

For  example:  Take  the  case  given  in  Art.  259,  where 
/  =  20,  m  =  7,  s  =  7,  n=\$,  r  =  13  and  v  =  3,  the  let- 
ters being  assigned  as  required  by  Fig.  55.  Here  the  tail 
beams  carried  by  the  header  at  A  are  7  feet  long,  and 
those  carried  by  the  two  other  headers  are  4  feet ;  there- 
fore A  =  7  and  B  —  C  —  4,  and  by  formulas  (100.)  and 

(101) 

ij . 

'  •  4x7  +  4x3)  =  45  •  85 


7  , .     4x13x3 

b    ==  ^(7X7 +  4X13)  +  — 2^--    =43-15 

The  result  here  obtained,  a/  being  larger  than  £',  shows 
that  the  case  has  been  rightly  assigned  to  the  first  class,  that 
of  Fig.  55. 


265.— Reassigning  the  Symbols. — The  result  of  a  compu- 
tation of  the  strains  may  show  that  the  arrangement  of  the 
symbols  was  erroneous  ;  instead  of  the  greatest  strain  being 
in  the  middle  it  may  be  found  at  one  side,  or  vice  versa. 
Then  the  lettering  of  the  loads  and  spaces  must  be  changed, 
to  agree  with  the  proper  diagram  and  formulas,  before  com- 
puting the  dimensions  of  the  beam  ;  using  formula  (96.)  or 
(97.)  for  the  class  shown  in  Fig.  55,  and  formula  (99.)  for  the 
class  shown  in  Fig.  56. 

266.— Example. — As  an  illustration  of  the  above,  take  a 
case  presumably  belonging  to  the  class  first  treated  (Fig.  55), 
where'  the  greatest  strain  is  an  outside  one.  Let  /  =  20 ; 
and  let  the  greatest  load,  1750  pounds,  be  designated  by 
A',  with  its  distances  in  =  7  and  ;/  =  13  ;  the  second 
load,  1250  pounds,  be  designated  by  B'y  with  its  distances 
r  =  12  and  J  =  8  ;  and  the  third  load,  1250  pounds,  be 
called  C,  and  its  distances  v  =  $  and  u  =  17.  To  find 


COMPOUND   LOADS — EXAMPLES.  2O/ 

the    united    effect   at    each    station,  we    have,  according   to 
formulas  (100.)  and  (101.), 

a'  =  —f  1750x13  +  1250x8  +  1250x3  \  —  12775 

8   / -\      1250x12x3 

b'=  --(1750x7  +  1250XI2J  +  -  —^-       =13150 

Here  b'  exceeds  a  and  shows  that  a  mistake  has 
been  made  as  to  the  class  to  which  the  case  belongs.  We 
must  change  the  symbols  and  arrange  them  for  the  second 
class  (Fig.  56). 

The  middle  weight  is  to  be  called  A' ;  the  weight  be- 
fore called  A',  at  7  feet  from  one  of  the  walls,  is  now  to 
be  B' ;  and  the  third  weight  C ' .  With  these  changes 
made,  we  have  ^'=1250,  £v  =  1750,  £' =  1250,  /  =  20, 
m  —  8,  Ji—12,  s  =  7,  r  —  13  and  ^  =  3;  and,  from  for- 
mulas (103.)  and  (104.), 

8   / \      1250x12x3 

a' =  —(1250x12  +  I750X7J  +  -  2_—       =13150 

b'  =  —(1250x8+  1750x13  +  1250x3^  =  12775 

The  result  is  now  satisfactory,  and  shows  that  the  prob- 
lem belongs  to  the  second  class,  the  one  in  which  the  great- 
est strain  occurs  at  the  middle,  and  this  notwithstanding  the 
fact  that  the  greatest  of  the  three  weights  is  at  the  outside. 
It  will  be  seen  that  the  results  of  the  two  trials  are  the  same, 
but  reversed,  that  which  was  at  first  taken  for  a'  being  now 
taken  for  b' . 

267.— Rule  for  Carriage  Beam  with  Three  Headers  and 
Two  Sets  of  Tail  Beams. — Formula  (99)  may  be  trans- 
formed so  as  to  make  it  specially  applicable  to  carriage 
beams. 

If,  in  Fig.s^y  we  suppose  the  spaces  EC  and  AB  to  be 
openings  in  the  floor,  then  one  set  of  tail  beams  will  extend 


208   COMPOUND  STRAINS,  GRAPHICALLY  EXPRESSED.    CHAP.  XII. 

s 

D 


from  C  to  A,  and  another  from  B  to  D,  giving-  three 
headers,  one  each  at  A,  B  and  C.  The  load  on  the 
header  A  will  equal  that  upon  C,  and  will  equal  one 
quarter  of  the  load  upon  the  space  occupied  by  the  tail 
beams  AC,  or  -±fg(in—ii).  Similarly  the  load  at  B 
will  be  \fgs.  Of  the  several  factors  composing  formula 
(99.)  we  now  have 


and  since 

and  the  formula  itself  becomes 


Cnv  =  \fgnv  (m—v) 
U—  \cfl 
n  =  \cfln 


b  =  jntfcfln  +  tfS*  ^^  +  i/^)  +  \fgnv(m-v)\ 


b  = 


t>  = 


b  = 


(cnl+gn  m—v  +g/)  +  \fgnv  (m—v]\ 


—v)  +gms*  +gnv(m—v)\ 


+gn(m—v)  ( 


b  =  - 


(106). 


CARRIAGE   BEAM   WITH   THREE   HEADERS.  209 

which  is  a  rule  for  carriage  beams  carrying  three  headers  and 
two  sets  of  tail  beams  relatively  placed  as  in  Fig.  56,  the  header 
producing  the  greatest  strain  being  between  the  other  two. 


268.—  Example.  —  What  should  be  the  width  of  a  carriage 
beam  20  feet  long,  12  inches  deep,  of  Georgia  pine  of 
average  quality,  carrying  three  headers  14  feet  long  ;  the 
headers  placed  so  as  to  afford  a  stair  opening  4  feet  wide 
at  one  wall,  and  a  light  well  5  feet  wide,  6  feet  from  the 
other  wall?  The  floor  beams  are  15  inches  from  centres 
and  carry  200  pounds  per  foot  superficial,  with  the  factor 
of  safety  a  =  4. 

In  this  case  we  have  B  =  850,  /=  200,  a  =  4,  c  =  ij, 
<^=  12,  /  =  2O,  v  =  4,  ;#  =  9,  •  #  =  11,  s  =  6,  r  =  14  and 
g  =  14,  and  by  formula  (106.) 


1  X2Q  +   4H  i4xu(9'-43)]  =  5-56 


or  the  breadth  should  be,  say    6    inches. 


210  COMPOUND  STRAINS,   GRAPHICALLY  EXPRESSED.   CHAP.  XII. 


QUESTIONS  FOR   PRACTICE. 


269. — In  a  beam  20  feet  long,  carrying  an  equably  dis- 
tributed load  of  2000  pounds,  and,  at  4  feet  from  one 
end,  a  concentrated  load  of  5000  pounds,  what  is  the  great- 
est strain  produced,  and  where  is  it  located  ? 

270. — In  a  floor  composed  of  beams  12  inches  deep, 
and  set  15  inches  from  centres,  there  is  a  Georgia  pine 
carriage  beam  22  feet  long,  carrying  two  headers  with  an 
opening  between  them.  The  headers  are  14  feet  long, 
and  are  placed  at  5  and  12  feet  respectively  from  the 
left-hand  wall.  The  floor  is  required  to  carry  200  pounds 
per  superficial  foot,  with  the  factor  of  safety  a  =  4. 

What  must  be  the  breadth  of  the  carriage  beam  ? 


CHAPTER  XIII. 

DEFLECTING     ENERGY. 

ART.   271. — Previously  Given  Rulc§  are  for  Rupture. — 

In  the  discussion  of  the  subject  of  transverse  strains,  the 
rules  adduced  thus  far  have  all  been  based  upon  the  resist- 
ance of  the  material  to  rupture,  or  the  power  of  the  material 
to  resist  the  destructive  effect  produced  by  the  load  which 
the  beam  is  required  to  carry. 

272. — Beam  not  only  to  Be  Safe,  but  to  Appear  Safe. — 

It  is  requisite  in  good  construction  that  a  loaded  beam  be 
not  only  safe,  but  that  it  also  appear  safe  ;  or,  that  the  amount 
of  deflection  shall  not  appear  to  be  excessive.  In  determining 
the  pressure  a  beam  may  receive  without  injury,  real  or  ap- 
parent, it  is  requisite  to  investigate  the  power  of  a  beam  to 
resist  bending,  rather  than  breaking — that  is,  to  ascertain  the 
Laws  of  Deflection. 

273. — All  Material  Possess  Elasticity. — Any  load,  how- 
ever small,  will  bend  a  beam.  If  the  load  be  not  excessive, 
the  beam  will,  upon  the  removal  of  the  load,  recover  its 
straightness. 

The  power  of  the  beam  by  which  it  returns  to  its  original 
shape  upon  the  removal  of  its  load,  is  due  to  the  elasticity 
of  the  material.  All  materials  possess  elasticity,  though 
some,  as  lead  and  clay,  have  but  little,  while  others,  as  india- 
rubber  and  whalebone,  have  a  large  measure  of  it. 


212  DEFLECTING   ENERGY.  CHAP.    XIII. 

274. — Limits  of  Elasticity  Defined. — When  a  beam  is 
bent,  some  of  its  fibres  are  extended  and  some  compressed, 
as  was  shown  at  Art.  22  ;  and  when  the  pressure  by  which 
the  bending  was  effected  is  removed,  the  fibres  resume  their 
original  length.  Should  the  pressure,  however,  have  been 
excessive,  then  the  resumption  will  not  be  complete,  but  the 
extended  fibres  will  remain  a  trifle  longer  than  they  were 
before  the  pressure,  and  the  compressed  fibres  a  trifle 
shorter.  When  this  occurs,  the  elasticity  is  said  to  be  in- 
jured ;  or,  the  pressure  has  exceeded  the  limits  of  elasticity. 

When  the  fibres  are  thus  injured,  they  are  not  only  in- 
capable of  recovering  their  original  length,  but  (the  pressure 
being  renewed  and  continued)  they  are  not  able  to  maintain 
even  their  present  length,  and  therefore  the  deflection  must 
gradually  increase,  and  the  fibres  continue  to  alter  in  length, 
until  finally  rupture  will  ensue. 

275. — A  Knowledge  of  the  Limits  of  Elasticity  Requisite. 

— To  secure  durability,  it  is  evident  that  a  beam  subject  to 
transverse  strain  should  not  be  loaded  beyond  its  limit  of 
elasticity.  Hence  the  desirability  of  ascertaining  this  limit. 

276. — Extension  Directly  as  the  Force. — Let  the  effect 
offeree  in  producing  extension  be  first  considered.  Suspend 
a  weight  of  one  pound,  by  a  strip  of  india-rubber  one  foot 
long,  and  measure  the  increase  in  the  length  of  the  rubber. 
Then,  double  the  weight,  and  it  will  be  found  that  the  in- 
crease in  length  will  be  double.  If  the  extension  caused  by 
one  pound  be  one  inch,  then  that  caused  by  two  pounds  will 
be  two  inches.  Three  pounds  will  increase  the  length  by 
three  inches  ;  or,  whatever  weight  be  suspended,  it  will  be 
found  that  the  extensions  will  be  directly  in  proportion  to 
the  forces  producing  them,  provided  always  that  the  force 


EXTENSION   DIRECTLY  AS   FORCE   AND    LENGTH. 


213 


applied  shall  not  be  so  great  as  to  destroy  the  elasticity  of 
the  material ;  shall  not  so  injure  it  as  to  prevent  it  from 
recovering  its  original  length  upon  the  removal  of  the 
force. 


277. — Extension  Directly  a§  the  Length. — The  above 
shows  the  relation  between  the  weight  and  the  extension. 
The  relation  will  now  be  shown  between  the  extension  and 
the  length  of  the  piece  extended.  At  5  inches  from  the 
upper  end  of  a  strip  of  rubber  attach  a  one-pound  weight. 
This  will  produce  an  extension  of,  say  a  quarter  of  an  inch. 
Detach  the  weight  and  re-attach  it  at  double  the  length,  or 
at  10  inches  from  the  upper  end.  It  will  now  be  found 
that  the  10  inches  has  become  ioj  inches;  the  elongation 
being  a  half  inch,  or  double  what  it  was  before.  Again, 
remove  the  weight  and  attach  it  at  15  inches  from  the 
upper  end,  and  the  strip  will  be  extended  to  I5f  inches; 
an  elongation  of  three  quarters  of  an  inch,  or  three  times  the 
amount  of  the  first  trial.  From  this  we  conclude  that,  under 
the  same  amount  of  pressure,  the  extensions  will  vary  directly 
as  the  lengths  of  the  pieces  extended. 


278. — Amount  of  Deflection. — When  the  projecting  beam 
ABCD,  Fig.  57,  is  deflected  by  a  .„ 
weight,  P,  suspended  from  the  | 
free  end,  it  bends  the  beam,  not  | 
only  at  the  point  A,  at  the  wall,  | 
but  also  at  every  point  of  its  length  | 
from  A  to  B,  so  that  the  line  | 
AB  becomes  a  convex  curve,  as  | 
shown. 

The  exact  shape  of  this  elastic 
curve  is  defined  by  writers  upon  that  subject.  A  full  discussion 


FIG.  57. 


214  DEFLECTING   ENERGY.  CHAP.  XIII. 

of  the  laws  of  deflection  would  include  the  development  of 
this  curve.  The  purpose  of  this  work,  however,  will  be  at- 
tained without  carrying  the  discussion  so  far.  All  that  will 
here  be  attempted  will  be  to  show  the  amount  of  deflection ; 
or,  in  the  present  example,  the  distance,  EB,  which  the 
point  B  is  depressed  from  its  original  position. 


279, — The  First  Step. — In  bending  a  beam,  the  fibres  at 
the  concave  side  are  shortened  and  those  at  the  convex  side 
are  lengthened.  The  first  step,  therefore,  in  finding  the 
amount  of  deflection,  will  be  to  ascertain  the  manner  of  this 
change  in  length  of  fibre,  and  the  method  by  which  the 
amount  of  alteration  may  be  measured. 


280. — Deflection  to  bo  Obtained  from  the  Extension. — 

It  is  manifest  that  the  elongation  of  the  fibres  in  the  upper 
edge  of  the  beam  AC,  Fig.  57,  must  occur  not  only  at  A, 
but  at  every  point  in  the  length  of  the  line  AB.  The  fibres 
at  every  point  suffer  an  exceedingly  small  elongation,  and  if 
we  can  determine  the  sum  of  this  large  number  of  small 
elongations,  we  shall  have  the  amount  of  extension  of  the 
line  AB.  This  may  be  done  in  a  simple  manner,  for  we 
may,  without  serious  error  in  the  result  to  be  obtained, 
consider  them  all  as  though  they  were  collected  and  concen- 
trated at  one  place  in  the  line,  instead  of  considering  each 
one  at  the  point  where  it  occurs. 

To  effect  this,  let  the  line  AB  be  drawn  straight,  as  in 
Fig.  58,  and  the  line  FG  be  drawn  at  right  angles  to  FK, 
the  neutral  line — the  line  which  divides  between  those 
fibres  which  are  extended  and  those  which  are  compressed, 


DEFLECTION   DIRECTLY   AS   EXTENSION. 


215 


and  therefore  a  line  in  which  the  fibres  are  not  altered 
in  length.  The  line  AG  maybe 
taken  as  the  sum  of  the  numerous 
small  extensions  which  have  oc- 
curred in  the  fibres  at  the  line  AB 
of  Fig.  57. 

In  order  to  show  the  relation 
between  the  extension  and  the  de- 
flection, we  will  investigate  the 
proportion  between  AG,  the  mea- 
sure of  the  one,  and  EB,  the  measure  of  the  other. 


281. — Deflection  Directly  a§  the  Extension. — Make  GJ, 
Fig.  58,  equal  to  AG,  and  draw  JL  parallel  with  AE. 
The  two  triangles  AGF  and  JGL  are  both  right-angled 
triangles,  and  if  AGF  be  revolved  ninety  degrees  upon  G 
as  a  centre,  then  the  line  AG  will  coincide  with  the  line 
GJ,  the  line  GF  with  the  line  GL,  and  AF  with  JL ; 
and  we  have  the  triangle  JGL,  equal  in  all  respects  to  the 
triangle  A  GF. 

The  triangle  GJL  is  homologous  with  the  triangle 
EBA,  for  the  right  line  AB  cuts  the  two  parallel  lines 
AE  and  JL,  making  the  angles  GLJ  and  EAB  equal; 
the  angles  at  E  and  G  are  by  construction  right  angles, 
and  hence  the  remaining  angles  at  J  and  B  must  be 
equal,  and  the  two  triangles,  having  all  their  respective 
angles  equal,  must  have  their  respective  sides  in  proportion, 
or  be  homologous.  Now,  since  the  triangle  JGL  is  iden- 
tical with  the  triangle  AGF,  we  have  the  two  triangles 
AGF  and  BEA  with  their  corresponding  sides  in  propor- 
tion, or 

GF-.AE-.AG'.EB 


and  as    AG    measures  the  extension  and   EB    the  deflection, 


2l6 


DEFLECTING    ENERGY. 


CHAP.    XIII. 


it  results  that  the  extension  is  in  direct  proportion  to  the 
deflection. 


282. — Deflection  Directly  as  the  Force,  and  a§  the 
Length. — By  the  experiment  of  Art.  276,  it  was  shown  that 
the  extensions  are  in  proportion  to  the  forces  producing 
them,  and  since,  as  just  shown,  they  are  also  in  proportion 
to  the  deflection,  therefore  the  deflections  are  in  direct  pro- 
portion to  the  forces  producing  them. 

In  the  case  of  a  semi-beam  pro- 
jecting from  a  wall,  as   AC,  Fig.  59, 
the  force  producing  the  deflection 
EBj    is  the  product  of  the  weight 
P,    into  the  arm  of  leverage    AE, 
at   the    end    of  which    the    weight 
acts  ;  or,  the   force   producing  the 
deflection  is  in  proportion  to  the 
weight  and  the  length. 
This  is  shown  in  Fig.  60.     Here  let  it  be  required  that  the 
weight    P   remain  constant  in  amount  and   location,  while 
the  length  of  the  semi-beam  be  increased.     We  shall  then 


FIG.  59. 


FIG.  60. 


have  at  E,  in  Fig.  60,  the  same  deflection  as  at  E  in  Fig.  59, 
because  the  force  producing  the  deflection  (PxAE)  is  the 
same  in  each  figure.  But  at  F,  the  end  of  the  increased 


DEFLECTION   DIRECTLY   AS   FORCE   AND    LENGTH.         2 1/ 

length,  the  deflection  is  greater,  owing  to  an  increase  in  the 
size  of  the  triangle  AEB,  from  AEB  to  AFC.  The  in- 
crease at  F  over  that  at  E  is  in  proportion  to  the  increase 
of  AF  over  AE,  because  EB  and  FC,  the  lines  measur- 
ing the  deflections,  are  similar  sides  of  the  two  homologous 
triangles  AEB  and  AFC  \  and  AE  and  AF,  the  lines 
measuring  the  lengths,  are  also  similar  sides  of  these  tri- 
angles. For  example,  if  AF  equal  twice  AE,  then  we  will 
have  FC  equal  to  twice  EB  ;  or,  in  whatever  proportion 
AF  is  to  AE,  we  shall  have  the  like  proportion  between 
FC  and  EB.  In  every  case,  the  deflections  will  be  in 
direct  proportion  to  the  lengths. 


283. — Deflection  Directly  a§  the  Length. —  Again:  If  the 
weight  be  moved  from  E  to  F,  Fig.  61,  the  end  of  the  above 
increased  length,  then  the  force  with  which  it  acts  is  in- 
creased, and  the  deflection  FC,  caused  by  the  weight  when 


FIG.  61. 

located  at  E,  now  becomes  FJ.  If  AF  equals  twice  AE, 
then  the  force  producing  deflection  is  doubled,  because  the 
leverage  at  which  the  weight  acts  is  doubled  ;  and  since 
the  deflections  are  in  proportion  to  the  forces  producing 
them,  FJ  is  double  FC\  and  in  whatever  proportion  the 
arm  of  leverage  be  increased,  it  will  be  found  that  the  deflec- 
tions at  the  two  locations  will  be  in  proportion  to  the  dis- 


218 


DEFLECTING   ENERGY. 


CHAP.    XIII. 


tances  of  the  weights  from  the  wall   AD,    or  in  proportion  to 
the  lengths. 

284. — Deflection  Directly  as  the  Length. — Once  more  : 
When  the  weight  was  located  at  E,  the  length  of  fibres  suf- 
fering extension  was  from  A  to  E,  but  now  this  length  is 
increased  to  AF. 

This  increase  in  length  of  fibres  will  increase  the  exten- 
sion (Art.  277),  and  consequently  the  deflection  (Art.  281). 
If  AF,  Fig.  62,  be  double  the  length  of  AE,  then,  owing  to 
the  extension  of  double  the  length  of  fibres,  the  deflection 
FJ,  Fig-  6l>  wiU  be  doubled,  or  increased  to  FK,  Fig.  62 ; 


FIG.  62. 

and  in  whatever  proportion  the  beam  be  lengthened,  the 
deflection  will  increase  in  like  proportion,  or  the  deflections 
will  be  in  proportion  to  the  lengths. 

285.— Total  Deflection  Directly  a§  the  Cube  of  the 
Length. — Summing  up  the  results  as  found  in  the  above 
several  steps  in  the  increase  of  deflection,  we  find,  by  a  com- 
parison of  Figs.  59  and  62,  that,  owing  to  an  increase  of  the 
beam  to  twice  its  original  length,  we  have  an  increase  in 
deflection  to  eight  times  its  original  amount.  If  EB  —  I, 


TOTAL   DEFLECTION   DIRECTLY   AS   CUBE   OF   LENGTH.    2 19 

then  FC=2,  F?=2FC=4,  and  FK=2F?=SEB.  With 
lengths  of  beam  in  proportion  as  i  to  2,  the  deflections  are 
as  i  to  8,  or  as  the  cubes  of  the  lengths. 

This  is  true  not  only  when  the  length  is  doubted,  but  also 
for  any  increase  of  length,  for  a  reference  to  the  discussion 
will  show  that  the  deflection  was  found  to  be  in  proportion 
to  the  length  on  three  several  considerations:  fast  (Art.  282), 
on  account  of  an  increase  in  the  size  of  the  triangle  contain- 
ing the  line  measuring  the  deflection  ;  second  (Art.  283), 
on  account  of  the  additional  energy  given  to  the  weight  by 
the  increase  of  the  leverage  with  which  it  acted  ;  and,  third 
(Art.  284),  on  account  of  the  extension  of  an  additional 
length  of  fibres.  The  deflection  and  the  length  being  neces- 
sarily of  the  same  denomination,  and  the  deflection  being 
taken  in  inches,  we  therefore  take  the  length,  N,  in  inches, 
and  we  have  the  deflection  in  proportion  to  NNN  or  to  Ns. 

286.— Deflecting  Energy  Directly  as  the  Weight  and 
Cube  of  the  Length. — From  Art.  276  the  extensions  are  in 
proportion  to  the  weights,  and  since,  from  Art.  281,  the  de- 
flections are  as  the  extensions,  therefore  we  have  the  deflec- 
tions in  proportion  to  the  weights.  Combining  this  with  the 
result  in  the  last  article,  we  have,  for  the  sum  of  the  effects, 
the  deflection  in  proportion  to  the  weight  and  the  cube  of 
the  length  ;  or, 

6  ;  PN* 


220  DEFLECTING   ENERGY.  CHAP.    XIII. 


QUESTIONS   FOR   PRACTICE. 


287. — The  rules  given  in  former  chapters  for  beams 
exposed  to  cross  strains  were  based  upon  the  power  of 
resistance  to  rupture. 

Upon  what  power  of  the  material  may  other  rules  be 
based  ? 

288. — To  what  degree  may  beams  be  deflected  without 
injury  ? 

289. — What  relation  exists  between  extensions  and  the 
forces  producing  them  ? 

290. — What  relation  exists  between  extensions  and 
deflections? 

291.— What  relation,  in  a  beam,  is  there  between  the 
deflections,  the  weights  and  the  lengths? 


CHAPTER  XIV. 

RESISTANCE     TO     FLEXURE. 

V 

ART.  292. — Re§istance  to  Rupture,  Directly  a§  the 
Square  of  the  Depth. — Having  considered,  in  the  last  chap- 
ter, the  power  exerted  by  a  weight  in  bending  a  beam,  atten- 
tion will  now  be  given  to  the  resistance  of  the  beam. 

It  was  shown  in  the  third  chapter,  that  the  resistance  to 
rupture  is  in  proportion  to  the  square  of  the  depth  of  the 
beam.  It  will  now  be  shown  that  the  resistance  to  bending 
is  in  proportion  to  the  cube  of  the  depth. 

293. — Resistance    to    Extension    Graphically    Shown. — 

For  the  greater  convenience  in  measuring  the  extension  of 
the  fibres  at  the  top  of  a  bent  lever  (Fig.  57),  it  was  proposed 
in  Art.  280  to  consider  this  extension  as  occurring  at  one 
point ;  at  the  wall.  In  an  investigation  of  the  resistance  to 
bending,  the  whole  extension  may  still  be  considered  as 
being  concentrated  at  that  point. 

Let  the  triangle  AGF,  Fig.  63,  represent  the  triangle 
AGF  of  Fig.  58,  in  which  AF  is  the  face  ot  the  wall,  and 
AG,  at  the  top  edge  of  the  lever,  is  the  measure  of  the  ex- 
tension of  the  fibres  there;  while  at  F,  the  location  of  the 
neutral  line,  the  fibres  are  not  extended  in  any  degree. 

It  is  evident  that  the  fibres  suffer  extension  in  proportion 
to  their  distance  from  F  towards  G,  so  that  the  lines  BCy 
DE,  etc.,  severally  measure  the  extensions  at  their  respec- 
tive locations.  Within  the  limits  of  elasticity,  the  resistance 


222 


RESISTANCE   TO    FLEXURE. 


CHAP.    XIV. 


FIG.  63. 


of  a  fibre  to  extension  is  measured  by  its  reaction  when  re- 
leased from  tension.  Thus,  the  line  BC  measures  the 
extension  of  the  fibres  at  that  location,  and  when  the  load  is 

removed  from  the  lever  these 
fibres  contract  and  resume 
their  original  length.  Hence, 
BC  also  measures  the  resist- 
ance to  extension.  The  resist- 
ance of  the  lever  to  bending, 
therefore,  is  in  proportion  to 
the  sum  of  the  extensions. 
The  extensions  of  that  por- 
tion of  the  lever  occurring  be- 
tween the  lines  A  G  and  BC 
is  measured  by  the  sum  of  the  lengths  of  all  the  fibres  within 
the  space  ABCG.  The  average  length  of  these  fibres  will 
be  that  of  the  one  at  the  middle,  and  the  number  of  fibres  is 
measured  by  CG,  the  width  they  occupy.  The  sum,  there- 
fore, of  the  lengths  of  all  the  fibres  will  be  equal  to  the  area 
of  the  figure  ABCG. 

Again,  the  sum  of  the  lengths  of  all  the  fibres  between 
the  lines  BC  and  DE  is  equal  to  the  area  of  the  figure 
BDEC;  so  in  each  of  the  other  figures  into  which  the 
triangle  AGF  is  divided  a  similar  result  is  found.  From 
this  we  conclude  that  the  sum  of  the  lengths  of  all  the  fibres 
exposed  to  tension  is  equal  to  the  area  of  the  whole  triangle 
AGF]  and,  therefore,  that  the  resistance  of  the  lever  is  in 
proportion  to  the  area  of  this  triangle. 


294. — Re§i§tance  to    Extension    in    Proportion    to    the 
Number  of  Fibrc§  and  their  Distance  from  Keutral  Line. — 

In  the  measure  of  the  extensions,  we  have  the  reaction  or 
power  of  resistance  ;  but  there  is  still  another  fact  connected 


RESISTANCE   OF   FIBRES   TO    EXTENSION.  223 

with  the  act  of  bending  which  needs  consideration.  The 
power  of  a  fibre  to  resist  deflection  will  be  in  proportion  to 
its  distance  from  F,  the  location  of  the  neutral  line  ;  or,  to 
the  leverage  with  which  it  acts,  as  was  shown  in  Figs.  8  and  9. 
Thus  at  AG  a  fibre  will  resist  more  than  one  at  DE,  while 
farther  down,  each  fibre  resists  less  until  at  F,  where  there 
is  no  leverage,  the  power  to  resist  entirely  disappears.  It 
may,  therefore,  be  concluded  that  the  power  of  each  fibre  to 
resist  is  in  proportion  to  its  distance  from  F ;  and  adding 
this  power  of  resistance  to  that  before  named,  we  have,  as 
the  total  resistance,  the  sum  of  the  products  of  the  lengths 
of  the  several  fibres  into  their  respective  distances  from  F. 


295. — Illustration. — As  an  illustration  of  the  above,  we 
may  find  an  approximate  result  thus  : 

Let  the  line  FG,  Fig.  63,  be  divided  into  any  number  of 
equal  parts,  and  through  these  points  of  division  draw  the 
lines  BC,  DE,  etc.,  parallel  with  AG.  These  lines  will 
divide  the  triangle  into  the  thin  slices  ABCG,  BDEC,  etc. 
Now,  the  resistance  of  the  top  slice,  ABCG,  will  be  approx- 
imately equal  to  its  area  into  its  distance  from  F\  or,  if 
CG,  the  thickness  of  the  slice,  be  represented  by  t,  and  the 
average  length  of  fibres  in  the  slice,  \(A  G  +  BC\  by  b,,  then 
the  area  of  the  slice  will  equal  btt ;  and,  if  at  be  put  for 
FG,  the  average  distance  of  the  slice  from  F  will  be 
at— \t\  and  therefore  the  resistance  of  the  top  slice  will  be 

R  =  *X«,-iO 

In  like  manner,  if  ct  be  put  for  the  average  length  of  the 
fibres  of  the  second  slice,  we  shall  have,  to  represent  its 
resistance, 


224  RESISTANCE   TO    FLEXURE.  CHAP.    XIV. 

For  the  third  we  shall  have 
R    = 


Thus,  obtaining  the  resistance  of  #//  the  slices  and  adding  the 
results,  we  have  the  total  resistance. 

296.  —  Summing  up  tlic  Re§istaiicc§    of  the  Fibre§.  —  To 

make  a  general  statement,  let  x  be  put  for  the  distance  from 
F  to  the  middle  of  the  thickness  of  any  one  of  the  slices  into 
which  the  triangle  is  divided,  and  let  r,  a  constant,  be  the 
length  of  an  ordinate,  as  DE,  located  at  the  distance  unity 
from  F.  Then  we  have  by  similar  triangles  the  proportion 

I    :    r    :  :    x    :    xr 

and  therefore  xr  will  equal  the  breadth  of  the  slice  at  any 
point  distant  x  from  F,  or  putting  x  equal  to  the  dis- 
tance from  F  to  the  middle  of  the  slice,  then  xr  will,  be 
equal  to  the  average  length  of  the  fibres  of  the  slice.  The 
resistance  then  of  one  of  the  slices,  say  the  top  slice,  will  be 
For  the  top  slice,  x=a,—%t,  therefore 


(al—  J/)V/  =  R 
Again;  for  the  second  slice,    x  —  at—%t     therefore 


For  the  third  slice  we  have 


In  like  manner  we  obtain  the  resistance  of  each  successive 
slice,  each  result  being  the  same  as  the  preceding  one,  ex- 
cepting the  fractional  coefficient  of  /,  which  differs  as  shown, 
the  numerator  increasing  by  the  constant  number  2.  When 


SUMMING  THE   RESISTANCES  OF  FIBRES.  225 

n    represents  the  total  number  of  slices,  then  the  last  result 
or  the  resistance  of  the  last  slice  will  be 


and  the  sum  of  all  the  resistances,  or 

R,  +  Ru  -f  Rni  +  etc.  +Rn  =  M 
will  equal  the  total  resistance  of  all  the  fibres,  thus  : 


M  =  (at—  \tjrt  +  (a—^tjrt  +  etc.  +  (a—  %2n—  itfrt 
M=  rt        -i/    +      - 


Now  the  number  of  slices  multiplied  by  the  thickness  of 

each  will  equal     FG,    or    nt  =  a  ,    from  which    /  =  —  ',     and, 

n 

by  substituting  this  value, 

X  /        i  \          2n—i 

a.—\t  —  a—\-  —  a  .(  i  —  —  1  =  a.  - 

'       n         '  \       2nj         '    2n 

and  (a,  -\tj  =  a^—^-  therefore 

4# 

M=rt  -V,^z3)!  +  etc.  +     ^=E 


M  =  --[(2«-  1)3  +  (2n—  3)2  +  etc.  4- 


Now,  (2«—  i)3  =  4^—  i  x 


(2n—  3)'  =  4«*—  3  x  4;*  +  9 
(2»—  5)a  =  4^—5  x  4^  +  25 

To  get  the  sum  of  these,  we  have,  first,  for  the  sum  of  the 
first  terms,     n  x  4n*  =  4^'. 


226  RESISTANCE   TO    FLEXURE.  CHAP.   XIV. 

The  coefficients  of  the  second  terms,  namely,  i,  3,  5, 
etc.,  equal  in  amount  the  sum  of  an  arithmetical  series  com- 
posed of  these  odd  numbers;  or,  n2  (Art.  200),  and  hence 
the  sum  of  these  several  second  terms  is  n2  x  ^n  =  4n3.  The 
first  and  second  terms  summing  up  alike  cancel  each  other, 
and  we  have  but  the  third  terms  remaining.  The  sum  of 
these  is  that  of  the  squares  of  the  odd  numbers  i,  3,  5,  etc., 
and  our  last  formula  becomes 

M  =  ^  [i2  4-  3*  +  52  +  etc.  +  (2n- 1)2] 

a.          ,     a*rt       afr  . 

Now,     /  =  -     and       '--?  —  -'— ,     therefore 


297. — True  Value  to  wliicli  these  Results  Approximate. 

— As  an  example  to  test  this  formula,  let     n  —  3,    then 


Again,  let     n  =  4,    then 


and  if    n  =  5,    then 


If    n  =  10,    then 


EXACT  AMOUNT   OF   RESISTANCE.  22? 

If    n  —  20,    then 


Reducing  these  five  fractions  to  their  least  common 
denominator,  43,200,  we  have 

When     n  =    3,  the  numerator  =  14,000 

n=    4,  "             "           =  H,i75 

n  =    5,  "             "           =  J4,256 

n  =  10,  "             "           =  14,364 

n  —  20,  "             "           =  14,391 

It  will  be  noticed  that  these  numerators  increase  as  n  in- 
creases, but  not  so  rapidly.  As  n  becomes  larger,  the  in- 
crease in  the  numerator  is  more  gradual,  but  still  remains 
an  increase,  for  however  large  n  becomes,  the  numerator 
will  still  increase,  until  n  becomes  infinite,  when  its  limit  is 
reached. 

This  limit  is  equal  in  this  particular  case  to  14,400,  or 
one  third  of  43,200,  the  denominator  ;  or,  in  general,  the 
value  of  the  fraction  tends  towards  ^  and 

M  =  %afr 

298.  —  True  Value  Defined  by  the  €alculu§.  —  This 
definite  result  is  reached  more  easily  and  directly  by  means 
of  the  calculus. 

Taking  the  notation  of  Art.  296,  we  have,  for  the  resist- 
ance of  one  of  the  slices,  the  expression 


This  gives  the  resistance  for  a  slice  at  any  distance,  x,  from 
F,  and  if  the  thickness  of  the  slice  be  reduced  to  the 
smallest  conceivable  dimension,  then  /,  its  thickness,  may 


228  RESISTANCE   TO    FLEXURE.  CHAP.    XIV. 

be  taken  for  the  differential  of   x,    or    dxy    and  we  have  as 
the  differential  of  the  resistance 

dR  =  x*rdx 
from  which,  by  integration,  is  obtained  (Art.  463) 


and  when  the  result  is  made  definite  by  taking  the  integral 
between  limits,  or  between    x  =  o     and     x=  at,    we  have 

R  =  ±a?r  (108.) 

299.  —  Sum  of  the  Two  Resistances,  to  Extension  and  to 
Compression.  —  The  foregoing  discussion  has  been  confined 
to  the  resistance  offered  by  that  portion  of  the  lever  the 
fibres  of  which  suffer  extension. 

A  similar  result  may  be  obtained  from  a  consideration  of 
the  resistance  offered  by  the  remaining  fibres  to  compression. 

If  c  be  put  to  represent  the  depth  of  that  part  of  the 
beam  in  which  the  fibres  are  compressed,  then  it  will  be 
found  that  the  resistance  to  compression  will,  from  (108.), 
be  equal  to 

R  =  tfr  (109.) 

and  the  total  resistance  offered  by  the  lever  will  be 


\ar  -f  tfr  = 

It  may  be  shown  also,  by  a  farther  investigation,  that  in 
levers  suffering  small  deflections,  or  when  not  deflected  be- 
yond the  limits  of  elasticity,  at  =  cy  or  the  neutral  line 
is  at  the  middle  of  the  depth.  In  the  latter  case,  we  have 
ttj  =  c  =  -J</,  and  therefore 


RESISTANCES   TO    EXTENSION   AND   COMPRESSION,   -     22Q 

300.  —  Formula  for  Deflection  in  Levers.  —  The  above 
is  the  result  in  a  lever  one  inch  broad.  A  lever  two  inches 
broad  would  bear  twice  as  much  ;  one  three  inches  broad 
would  bear  three  times  as  much  ;  or,  generally,  the  resistance 
will  be  in  proportion  to  the  breadth.  We  have  then  for  a 
lever  of  any  breadth 

(110.) 


This  expression  gives  the  resistance  to  the  deflecting 
energy,  which  is  (Art.  286)  equal  to  PN3.  This  power, 
PN3,  however,  not  only  overcomes  the  resistance,  ^rbd3, 
but  in  the  act  also  accomplishes  the  deflection  ;  moves  the 
lever  through  a  certain  distance.  Representing  this  distance 
by  eJ,  we  have,  as  the  full  measure  of  the  work  accom- 
plished, d  x  -^rbd3.  When  the  power  and  the  work  are 
equal,  we  have 

PN3  =  -i^rbd3  from  which, 

PN3 


301.  —  Formula  for  Deflection  in  Beams.  —  The  expression 
(111.)  is  for  a  semi-beam  or  lever.  When  a  full  beam,  sup- 
ported at  each  end,  is  deflected  by  W,  a  weight  located 
at  the  middle,  we  have  to  consider  that  for  P  we  must 
take  %W,  and  for  N  take  \L  (see  Art.  35).  These 
alterations  will  produce 


WL* 

= 


for  the  deflection  of  a  beam  supported  at   both   ends   and 
loaded  in  the  middle. 


230  RESISTANCE   TO    FLEXURE.  CHAP.   XIV. 

302.  —  Value  of  F,  the  Symbol  for  Resistance  to  Flexure. 

—  In  formula  (112.)  the  dimensions  are  all  in  inches.  As  it  is 
more  convenient  that  .the  length  be  taken  in  feet,  let  / 
represent  the  length  in  feet,  then 

y^=/,       L=i2l    and     L3  =  ~i2l3  =  172*1* 
By  substitution  in  formula  (112.)  we  have 


1728  J^75  _  1296*07* 
~  rbd3 


Wl 


1296        M'd 

The  symbol     r     is  a  measure   of  the  extension,  differing  in 
different    materials,    but    constant,    or   nearly    so,    in    each. 

Putting  for   -  ^   the  letter    F    we  have 


The  respective  values  of  F  for  several  materials  have 
been  obtained  by  experiment,  and  may  be  found  in  Table 
XX.  Its  value  in  each  case  is  that  for  a  beam  supported  at 
each  end,  and  with  the  load  in  pounds  applied  at  the  middle 
of  /,  the  distance  in  feet  between  the  bearings  ;  while  b, 
d  and  6  are  in  inches  ;  6  being  the  deflection  within 
the  limits  of  elasticity. 

303.—  Conipari§on  of  F  with  E9  the  Modulus  of  Elas- 
ticity. —  The  common  expression  for  flexure  of  beams  when 
laid  on  two  supports  and  loaded  at  the  middle  is  [Tate's 
Strength  of  Materials,  London,  1850,  p.  24,  formula  (49*)] 


COEFFICIENT   OF   ELASTICITY.  231 

in  which  E  represents  what  is  termed  the  Modulus  or  Co- 
efficient of  Elasticity,  which  is  (same  work,  p.  3),  "  that  force, 
which  is  necessary  to  elongate  a  uniform  bar,  one  square 
inch  section,  to  double  its  length  (supposing  such  a  thing 
possible)  or  to  compress  it  to  one  half  its  length"  ;  and  / 
represents  the  Moment  of  Inertia  (Arts.  457  to  4-63)  of  the 
cross-section  of  the  beam. 

In  this  expression  the  dimensions  are  all  in  inches.     To 

change     L    to  feet  we  have    — =  /  equals  the  length  in  feet, 

or  L3=  i27*=  i;28/5. 

Substituting  this  value  in  (114-}  we  obtain 

1728  Wl3       $6Wl3 

6  —  — 


or  _ 

36 

In  formula  (113.}  we  have 

F^-1 
Multiplying  this  by    12    gives 


48^7  El 

E        Wls 


Wl 


and  since     -fabd*  —  I    (see  Art.  463) 


Comparing  this  with  above  value  of    —^   we  have 

jg 

--  =  \2F    or    E— 


232  RESISTANCE   TO    FLEXURE.  CHAP.   XIV. 

304.— Relative  Value  of  F  and  E.— In  Table  XX.  the 
value  of  F  for  wrought  iron  is,  from  experiments  on  rolled 
iron  beams,  62,000.  Then 

432^  =  E  —  432  x  62000  =  26784000, 

cquab  the  modulus  of  elasticity  for  the  wrought-iron 
of  which  these  beams  were  made.  They  were  of  Ameri- 
can metal.  Tredgold  found  the  value  for  English  iron 
to  be  £  =  24,900,000;  and  Hodgkinson  from  19,000,000  to 
28,000,000. 

An  average  of  the  results  in  seven  cases  gives  25,300,000 
as  the  modulus  of  elasticity  for  English  wrought-iron. 


305.  —  Comparison    of    F    with     E    common,    and    with 

the    E  of  Barlow.—  Barlow,   in    his    "  Materials   and    Con- 
struction/' p.  93,  foot-note  (Ed.  of  1851),  uses  the  expression 

L3  W  L3  W 

-T-TS*  =  E,    instead  of    —  TT^  >    f°r  a  lever  loaded  at  one  end  ; 

UW  ' 

and  on  p.  94,      -  f-hiiT  =  £•     1  ne  dimensions  are  all  in  inches. 


Changing  the  length  to  feet,  we  have 


bd't 

E       wr 


108 


Comparing  this  with  (113.),  which  is 


ld'6 


we  have   -    '$-  =  F,     or     ioSF=E.     We  found   before  (Art. 

IOo 

303)  that     E  =  432^,     and   since     4  x  108  =  432,    therefore 


COMPARISON   OF   CONSTANTS.  233 

the  E  of  Barlow  equals  one  quarter  of  the  E  in  common 
use,  and  his  values  of  E  are  equal  to  108  times  the  values 
of  F  as  given  in  this  book. 

For  example  ;  on  p.  147,  in  an  experiment  on  New  Eng- 
land fir,  he  gives,  by  an  error  in  computation,  E  =  54780x5, 
but  which,  corrected,  equals  373326.  Dividing  this  by  108 
as  above,  gives 


By  reference  to  Table  XX.  we  find  that  for  spruce,  the 
wood  most  probably  intended  for  New  England  fir, 

F  =  3500 

Again;  taking  Barlow's  four  experiments  on  oak,  p.  146, 
and  correcting  the  arithmetical  errors,  we  have  E  =  361758, 
482344,  291227  and  242860.  This  gives  an  average  of 
344547,  and  dividing  it  by  108  as  above,  we  have 

F=  3190 

By  reference  to  Table  XX.  we  find  that  by  my  experi- 
ments 

F  =  3100 


306.— Example  under  the  Rule  for  Flexure. — To  make 
a  practical  application  of  the  rule  in  formula  (113.\  let  it  be 
required  to  find  the  depth  of  a  white  pine  beam  10  feet 
long  between  bearings  and  4  inches  broad;  and  which,  with 
a  load  of  2000  pounds  at  the  middle  of  it^length,  shall  be 
deflected  0-3  of  an  inch. 


234  RESISTANCE  TO   FLEXURE.  CHAP.   XIV. 

We  obtain  from  (113.) 


or,  in  this  case, 


Wl3 

d3  —  — - 
~  Fbd 


d3  —     200°  x  IC)8 
2900  x  4  x  o  •  3 

2000000 


or  the  depth  should  be     8-31,     say     8£    inches. 


QUESTIONS   FOR   PRACTICE. 


307. — How  may  the  resistance  of  a  fibre  to  extension  be 
measured  when  the  elasticity  remains  uninjured  ? 

308. — In  a  beam  exposed  to  transverse  strain,  what  is  the 
resistance  to  extension  in  proportion  to  ? 

309. — When  the  bending  energy  and  the  resistance  of  a 
beam  are  in  equilibrium,  what  is  the  expression  for  this 
relation  ? 

310. — Given  a  white  pine  beam  20  feet  long,  6  inches 
broad  arid  12  inches  deep,  and  loaded  with  1000  pounds 
at  the  middle.  What  will  be  the  deflection,  the  value  of  '  F 
being  2900  ? 


CHAPTER  XV. 

RESISTANCE   TO   FLEXURE— LIMIT   OF   ELASTICITY. 

ART.  311. — Rule§  for  Rupture  and  for  Flexure  Com- 
pared.— The  rules  for  determining  the  strength  of  materials 
differ  from  those  denoting  their  stiffness.  The  former  are 
more  simple  ;  all  their  symbols  being  unaffected  except  one, 
and  this  only  to  the  second  power,  or  square ;  in  the  latter, 
two  of  the  symbols  are  involved  to  the  third  power  or  cube. 

Many,  in  determining  the  dimensions  of  timbers  exposed 
to  transverse  strains,  are  induced,  by  the  greater  simplicity 
of  the  rules  for  strength,  to  use  them  in  preference  to  those 
for  stiffness,  even  when  the  latter  only  should  be  used. 

A  beam  apportioned  by  the  rules  for  strength  will  not 
bend  so  as  to  strain  the  fibres  beyond  their  elastic  limit,  and 
will  therefore  be  safe  ;  but  in  many  cases  the  beam  will  bend 
more  than  a  due  regard  for  appearance  will  justify. 

When  timbers,  therefore,  as  those  in  the  ceiling  or  floor 
of  a  room,  might  deflect  so  much  as  to  be  readily  percep- 
tible, and  unpleasant  to  the  eye,  they  should  have  their 
dimensions  fixed  by  the  rules  for  stiffness  only. 


312. — The  Value  of  a,  the  Symbol  for  Safe  Weight. — In 

order  that  the  symbol  a  in  the  rules  for  strength,  denoting 
the  number  of  times  the  safe  weight  is  contained  in  the 
breaking  weight,  may  be  of  the  proper  value  to  preserve  the 
fibres  of  the  timber  from  being  strained  beyond  the  elastic 


236  RESISTANCE  TO  FLEXURE— LIMIT  OF  ELASTICITY.  CHAP.  XV. 


limit,  a  few  considerations  will  now  be  presented  showing  the 
manner  in  which  this  value  is  ascertained. 

In  T^.  64  let  ABCD  represent  a  lever  with  one  end,  AD, 
imbedded  in  a  wall,  AD  being  the  face  of  the  wall,  and  car- 
rying at  the  other  end,  BC,  a 
weight  P\  the  weight  de- 
flecting the  lever  from  the  line 
AE  to  the  extent  EB.  The 
line  FH  is  the  neutral  line, 
and  FG  is  drawn  at  right 
angles  to  FH. 

As  in  Figs.  58  and  63,  so 
here  the  triangle  AFG  shows 
the  elongation  of  the  fibres  in 
the  upper  half  of  the  beam, 
and  AG  the  elongation  to  the  limits  of  elasticity  of  the 
fibres  at  the  upper  edge  AB.  The  triangle  AFG  is  in  pro- 
portion to  the  triangle  ABE,  as  shown  in  Art.  281.  If 
AB=N  (this  being  a  semi-beam),  and  e  equals  the  exten- 
sion per  unit  of  N,  then  A  G  =  eN. 
We  have  by  similar  triangles 

AF  :  AB  ::  AG  :  EB 
Then  if    AD  =  d    and    EB  =  d 


\d  :  N  :  :  eN  :  6  = 
2eN* 


FIG.  64. 


The  dimensions  here  are  all  in  inches.     To  change  N  in 
inches  to   n   in  feet,  we  have 


N 

—  =  n,    N—lIn    and     </V*  =  144;** 


MEASURE   OF   EXTENSION   OF   FIBRES.  237 

from  which 


and  from  this  we  obtain 

dd 


e  = 


288^ 


in  which  d  is  the  deflection  when  at  the  limit  of  elasticity, 
and  in  which  e,  d  and  <?  are  in  inches,  and  n  in  feet. 
This  is  for  a  semi-beam,  and  it  will  be  perceived  that  the 
deflection  EB,  in  Fig.  64,  caused  by  the  weight  P,  is  pre- 
cisely the  same  as  would  be  produced  in  a  full  beam  by  dou- 
ble this  weight  placed  at  Z>,  the  beam  being  in  a  reversed 
position. 

When,  therefore,  /  equals  the  length  of  the  full  beam  in 
feet,  n  will  equal  \l .  Substituting  this  value  of  n  in  the 
above  expressions,  we  have 


d  (116> 

and  for  the  value  of   e, 

dd 


In  Art.   302   we   have,  for    the    stiffness^  of    materials, 
formula  (113.), 

F^W* 


For  (5  substitute   *-j — ,   its  value  as  just  found,  and,  in 

order  to  distinguish  the  weight  used  to  produce  flexure 
from  that  used  to  produce  rupture,  let  us  for  the  moment 
indicate  the  former  by  £,  and  the  latter  by  W.  Then, 


238  RESISTANCE  TO  FLEXURE—  LIMIT  OF  ELASTICITY.  CHAP.  XV. 

from  the  above, 


Gl3  = 

a 

Gl  —  j2Fbd2e 

The  relation  between  F,  the  measure  of  the  elasticity  of 
materials,  and  £>,  the  resistance  to  rupture,  may  be  put 
thus  : 

F 
B  :  F  :  :    i  :  m  =  -^  ;    or,     F  =  Bm 

£> 

Substituting  this  value  for    F    in  the  above,  we  have 

Gl  =  J2Bmbd*e 
Gl 


=  Bbd* 


Wl 
Now  the  formula  lor  strength,    B  —  ,-j^,      ^form.  (10.)  in 

Art.  36]  gives     Wl  —  Bbds ;     a  comparison  of  this  value  of 
Bbds   with  that  above  shown  gives 

Gl 


7  2  em 


=  Wl 


Since  G  is  the  deflecting  weight  which  bends  the  lever 
to  the  limit  of  elasticity,  it  is  therefore  the  ultimate  weight 
which  may  be  trusted  safely  upon  the  beam,  and  as  a  is  a 
symbol  put  to  denote  the  number  of  times  G  is  contained  in 
Wy  the  breaking  weight,  therefore 

W 
G   :    W  :  :   I  :  a  =  ~     and     Ga  =  W 

(jr 

Substituting  this  value  for     W    in  the  above,  we  have 

Gl 

==  Gal 


~ 


MEASURE   OF   SYMBOL   FOR   SAFETY.  239 

p 

As  above  found,   m  =  -5- ,  therefore 

Z) 


From  this  expression  the  values  of  a  for  various  ma- 
terials have  been  computed,  and  the  results  are  to  be  found 
in  Table  XX. 


313. — Rate  of  Deflection   per  Foot  Length  of  Beam. — 

The  value  of   a    as  just  found  is  based  upon  the  elasticity  of 
the  material,  and  is  measured  by  this  elasticity  at  its  limit. 

This  limit  is  that  to  which  bending  is  allowable  in  beams 
apportioned  for  strength.  In  beams  required  to  sustain  their 
loads  without  bending  so  much  as  to  be  perceptible  or  offen- 
sive to  the  eye,  the  bending  is  generally  far  within  the  elastic 
limit.  The  deflection  in  these  beams  is  rated  in  proportion 
to  the  length  of  the  beam  ;  or,  when  r  in  inches  equals  the 
rate  of  deflection  per  foot  in  length  of  the  beam,  then  rl=  rf. 
The  deflection  by  formula  (116.)  is 


d 
therefore  ri  — 


r  = 


This  gives  r  at  its  greatest  possible  value,  and  shows 
that  it  should  never  exceed  72  times  the  ratio  between  the 
length  and  depth,  multiplied  by  e ;  e  being  the  measure 
of  extension  as  recorded  in  Table  XX.  The  ratio  between 


240  RESISTANCE  TO  FLEXURE — LIMIT  OF  ELASTICITY.  CHAP.  XV. 

the  length  and  depth  is  to  be  taken  with  /  in  feet  and  d 
in  inches. 

The  value  of  r  as  required  in  beams  of  the  usual  pro- 
portions and  deflection,  will  not  be  as  great  as  that  here  shown 
to  be  allowable.  In  cases  where  the  rate  of  deflection,  r,  is 
as  great  as  0-05  of  an  inch  per  foot,  and  the  length  of  the 

beam  is  short  in  comparison  with  the  depth  (say  -j-  is  as 
small  as  -  V  then  there  will  be  danger  of  r  exceeding  the 
limit  fixed  by  this  rule.  When  the  fraction  -7-  is  less  than 

-   then  the  rate    r    should  be  tested  to  know  whether  it  has 

exceeded  the  proper  limit.  It  is  seldom,  however,  that  a 
beam  7  inches  high  is  used  shorter  than  5  feet,  or  one 
14  ,  inches  high  shorter  than  10  feet.  Generally  the  num- 
ber of  feet  in  the  length  exceeds  the  number  of  inches  in  the 
depth. 

3(4.— Rate  of  Deflection  in  Floors. — The  rate  of  deflec- 
tion allowable  so  as  not  to  be  unsightly  is  a  matter  of  judg- 
ment. Tredgold,  in  his  rules  for  floor  beams,  fixed  it  at  ^ 
of  an  inch  per  foot  of  the  length,  or  0-025.  This  is  thought 
by  some  to  be  rather  small,  especially  since  in  floors  the  limit 
of  the  rate  is  seldom  reached  ;  in  fact  never,  except  when  the 
floor  is  loaded  to  its  fullest  capacity,  a  circumstance  which 
occurs  but  seldom,  and  then  only  for  a  limited  period.  For 
this  reason,  it  is  proper  to  fix  the  rate  at  say  -£s,  or  0-03 
of  an  inch  per  foot.  With  this  as  the  rate  for  a  full  load,  the 
usual  rate  of  deflection  under  ordinary  loads  will  probably 
not  exceed  o-oi  or  0-015.  In  the  rules,  the  symbol  r 
is  left  undetermined,  so  that  the  rate  may  be  fixed  as  judg- 
ment or  circumstances  may  dictate  in  each  special  case. 


QUESTIONS   FOR    PRACTICE. 


315. — What    is   the    distinction    between   the   rules  for 
strength  and  those  for  stiffness! 

316. — What  expression   shows    <?,     the  deflection  at  the 
elastic  limit? 

317. — What  expression  gives  the  measure  of  extension  at 
the  elastic  limit  ? 

318. — What  expression  shows   the  ultimate  value  of    a, 
the  factor  of  safety  ? 

319. — What  expression   gives  the   ultimate  value  of    r, 
the  rate  of  deflection  ? 


CHAPTER   XVI. 

RESISTANCE  TO   FLEXURE— RULES. 

ART.  320. — Deflection  of  a  Beam,  with  Example. — The 

formula  (113.)  for  the  deflection  of  beams  supported  at  each 
end  and  loaded  at  the  middle,  is 

E*  __ . 

from  which,  d  =  -^7-75  (120.) 


This  is  the  deflection  of  any  beam  placed  and  loaded  as 
above.  For  example:  \Vhatisthedeflectionofawhitepine 
beam  of  4  x  9  inches,  set  edgewise  upon  bearings  16  feet 
apart,  and  loaded  with  5000  pounds  at  the  middle  ;  the 
value  of  F  being  2900,  the  average  of  experiments,  the 
results  of  which  are  recorded  in  Table  XX.  ? 

The  deflection  in  this  case  will  be 

5000  x  i63         50  x  1024 

=  2-4218 


3 
2900  x  4  x  9         29  x  729 

This  is  a  large  deflection,  much  beyond  what  would  be 
proper  in  a  good  floor,  for  at  0-03  inch  per  foot  of  the 
length  of  the  beam,  the  rate  of  deflection  adopted  (Art.  314), 
we  should  have 

6  =  16  x  0-03  ==  0-48 

or,  say  half  an  inch,   whereas  the    5000    pounds  upon  this 


PRECAUTIONS   IN   REGARD   TO   CONSTANTS.  243 

beam  produces  five  times  this  amount.  Although  so  greatly 
in  excess  of  what  a  respect  for  appearance  will  allow,  it  is 
still,  however,  within  the  limits  of  elasticity,  as  will  be  seen 
by  the  use  of  formula  (116.),  in  which  we  have 


Obtaining  from  Table  XX.  the  average  value  of  e,  equal 
0-0014,   we  have 

72  x  0-0014  x  i6a 
d  =  -  -  =  S  x  0-0014x256  =  2-8672 

as  the  greatest  deflection  allowable. 


321. — Precautions  as  to  Values  of  Constants  F  and  e. — - 

The  above  is  the  ultimate  deflection  within  the  limits  of 
elasticity,  and  is  0-4454  in  excess  of  the  2-4218  produced 
by  the  5000  pounds.  In  general,  it  would  be  undesirable  to 
load  a  beam  so  heavily  as  this,  or  to  deflect  it  to  a  point  so 
near  the  limit  of  elasticity,  and,  unless  the  timber  be  of  fair 
quality,  would  hardly  be  safe. 

Some  pine  timber  would  be  deflected  by  this  weight  much 
more  than  is  here  shown — in  fact,  beyond  the  limits  of  elas- 
ticity. In  the  above  computation,  F  was  taken  at  2900, 
the  average  value,  and  the  measure  of  elasticity,  e,  was 
taken  at  0-0014,  also  the  average  value  ;  whereas,  had  these 
constants  been  taken  at  their  lowest  value,  such  as  pertain  to 
the  poorer  qualities  of  white  pine,  and  in  which  F=  2000 
and  ^  =  0-001016,  the  limits  of  elasticity  would  have  been 
found  at  a  trifle  over  2  inches,  while  the  deflection  would 
have  reached  3^  inches. 


244  RESISTANCE  TO   FLEXURE  —  RULES.        CHAP.  XVI. 

322.  —  Values  of  Constants  J^  and  c  to  foe  Derived  from 
Aetual  Experiment  in  Certain  Cases.  —  For  any  important 
work,  the  capacity  of  the  timber  selected  for  use  should  be 
tested  by  actual  experiment.  This  may  be  done  by  submit- 
ting several  pieces  to  the  test  of  known  weights  placed  at  the 
centre,  by  increasing  the  weights  by  equal  increments,  and 
by  noting  the  corresponding  deflections.  From  these  deflec- 
tions, the  specific  values  of  F  and  e  for  that  timber  may  be 
ascertained  ;  and  with  these  values  the  timber  may  be 
loaded  with  certainty  as  to  the  result.  In  the  absence  of  a 
knowledge  of  the  elastic  power  of  the  particular  material  to 
be  used,  a  sufficiently  wide  margin  should  be  allowed,  in 
order  that  the  timber  may  not  be  loaded  beyond  what  the 
poorer  kinds  would  be  able  to  carry  safely. 

323.  —  Deflection  of  a  L,ever.  —  The  rule  for  deflection,  as 
discussed  in  these  last  articles,  is  appropriate  for  a  beam  sup- 
ported at  both  ends  and  loaded  in  the  middle.  A  rule  will 
now  be  developed  for  a  semi-beam  or  lever;  a  timber  fixed 
at  one  end  in  a  wall,,  and  with  a  weight  suspended  from  the 
other.  The  deflection  in  this  case  is  precisely  the  same  as 
that  produced  by  twice  the  weight,  laid  at  the  middle  of  a 
whole  beam,  double  the  length  of  the  lever,  and  supported  at 
each  end. 

Let  the  weight  at  the  end  of  the  lever  be  represented  by 
P,  and  the  length  of  the  lever  by  n,  then  W  of  formula 

wr 

(120),  which  is    d  =  -pi~r3  »  **i«  equal   2P,   and   /  will  equal 
2-n,   and  we  have,  by  substituting  these  values  for    W  and   / 


Fbd 


DEFLECTION   WITHIN   THE   ELASTIC    LIMIT.  245 

324. — Example. — The  deflection  above  found  is  that  pro- 
duced in  a  lever  by  a  weight  suspended  from  its  free  end. 

As  an  example :  What  would  be  the  deflection  caused  by 
a  weight  of  1500  pounds  suspended  from  the  free  end  of  a 
lever  of  Georgia  pine,  of  average  quality,  3x6  inches 
square  and  5  feet  long  ? 

Here  we  have  P=  1500,  n=  $,  F=  5900,  b  =  3  and 
d  =  6 ;  and  therefore 

16  x  1500  x  53       80  x  125 
6—-  —4-  =  -  ^  =  0-7847 

5900  x  3  x  6        59  x  216 


325. — Te§t  by  Rule  for  Elastic   Limit  in   a    Lever. — To 

test  the  above,  to  ascertain  as  to  whether  the  deflection  is 
within  the  limits  of  elasticity,  take  /  =  2n  =  10,  and  by 
formula  (116.)  we  get 

72d2         72  X  0-00109  X  I02 
d  =  -L—r—  =  L-        ~~g~~  -=12x0-109=1.308 

This  is  satisfactory,  as  it  shows  that  the  lever  has  a  de- 
flection (0-7847)  of  not  much  more  than  half  that  within  the 
elastic  limit  (i  -308),  and  therefore  a  safe  one. 


326. — Load  Producing  a  Given  Deflection  in  a  Beam. — 

By  inversions  of  formulas  (120.)  and  (121),  we  may  have 
rules  for  ascertaining  the  weight  which  any  beam  or  level 
will  carry  with  a  given  deflection. 

First ;  for  a  beam,  we  take  formula  (120.) 

Wl3 
~~Fbds 

and  have 


246  RESISTANCE   TO    FLEXURE — RULES.          CHAP.  XVI. 

327.— Example. — Foran  example:  What  weight  upon  the 
middle  of  a  beam  of  spruce,  of  average  quality,  5  inches 
broad,  10  inches  high,  and  20  feet  long  between  the  bear- 
ings, will  produce  a  deflection  of  0-03  inch  per  foot,  or 
0-6  inch  in  all? 

Here  we  have  F=  3500,  £  =  5,  d—  10,  6  =  0-6  and 
/—  20;  therefore 

3500  x  5  x  io3  x  0-6       10500 

W.**.r       -53-       -  =  -g— =1312.5 


328. — Load  at  the  Limit  of  Elasticity  in  a  Beam. — Again : 
What  weight  could  be  carried  upon  this  beam  if  the  deflec- 
tion! were  permitted  to  extend  to  the  limit  of  elasticity  ? 

Formula  (116.)  gives  us 


and  from  Table  XX.  we  have  the  average  value  of   e    for 
spruce  equal  to   0-00098,  and  therefore 

72  x  0-00098  x  202 
d=  ~  -^  72  x  0-00098  x  40=  2-8224 


Substituting  this  new  deflection  in  the  former  statement, 
we  have 


W=  25«  UJIJ^.8224  =  49_39£  =  ^ 

This  6174  pounds  for  good  timber  would  be  a  safe  load, 
but  if  there  be  doubts  as  to  the  quality,  the  load  should  be 
made  less  according  to  the  lower  Values  of  F  and  e. 


DEFLECTION   AT   THE   ELASTIC    LIMIT.  *        247 

329.  —  Load   Producing  a  Given  Deflection  in  a  Lever- 
Example.  —  Second  ;  for  a  lever,  we  take  formula 

i6/V 
6= 
and  find  by  inversion 


. 

An  application  of  this  rule  may  be  shown  in  the  answer 
to  the  question  :  What  weight  may  be  sustained  at  the  end  of 
a  hemlock  lever,  6  inches  broad  and  9  inches  high,  firmly 
imbedded  in  a  wall,  and  projecting  8  feet  from  its  face  ? 
The  hemlock  is  of  good  quality,  and  the  deflection  is  limited 
to  i  inch. 

Here  we  have  ^=2800,  b  =  6,  d  =  9,  d=i,  and  n  =  8  ; 
therefore 

_  2800  x  6  x  93  x  i  _ 
~H6~^V~ 

that  is,    1495    pounds  at  the  end  of  the  lever  would  deflect  it 
one  inch. 


330. — Deflection  in  a  Lever  at  the  Limit  of  Elasticity. — 

What  deflection  in  this  lever  would  mark  the  limit  of  elas- 
ticity ? 

Formula  (110.)  is 


Taking    /    at   twice     n     we  have     /=:  16,     ^==9,     and 
—  0-00095 ;    and  as  a  result 

72x0-00095  x  i62 
os=v~ — ^  -==8x0.00095x256=1.9456 


248       .  RESISTANCE   TO    FLEXURE  —  RULES.         CHAP.  XVI. 

331.  —  Load  on  Lever  at  the  Limit  of  Elasticity.  —  What 
weight  would  deflect  this  lever  to  the  limit  of  elasticity  ? 
For  this  we  have 


2800  x  6  x  o3  x  i  -9456 

p=~  - 


This  is  nearly  double  the  weight  required  to  deflect  it  one 
inch,  as  before  found  ;  and  the  deflection  is  also  nearly 
double.  The  weight  and  the  deflection  are  directly  in  pro- 
portion. If  1500  pounds  deflect  a  beam  one  inch,  3000 
pounds  will  deflect  it  two  inches. 


332. — Value§   of   W,  I,  6,  d   and   6  in   a   Beam. — By   a 

proper  inversion  of  the  formulas  for  beams,  any  one  of  the 
dimensions  may  be  obtained,  provided  the  other  dimensions 
and  the  weight  are  known. 
Thus  we  have  (form. 


3 


and  from  this  find 

the  length, 

the  breadth,          b  =  ^  (125) 

and  the  depth,      d  = 

and,  as  in  formula  (120)t 

Wl9 

the  deflection,       <J  =  -757-73 
rba 


DEFLECTION  —  DIMENSIONS   OF   BEAM.  249 

333.  —  Example—  Value  of  I  in  a  Beam.  —  Take  an  exam- 
ple under  formula  (124-}-  What  should  be  the  length  of  a 
beam  of  locust  of  average  quality,  4  inches  broad  and  8 
inches  high,  to  carry  5000  pounds  at  the  middle,  with  a 
deflection  of  one  inch  ? 

In  formula   (124}    ^=5050,    £  =  4,    d=%,    6  =  i    and 

W=  5000;  hence  ___ 

x  4  x  83  x  i 

=12-74 


5000 
or  the  answer  is  \2\  feet. 

334.  —  Example—  Value  of  b  in  a  Beam.  —  As  an  example 
under  formula  (125.},  let  it  be  required  to  know  the  proper 
breadth  of  an  oak  beam  of  average  quality.  The  depth  is 
6  inches  and  the  length  10  feet.  The  load  to  be  carried  is 
500  pounds  placed  at  the  middle,  and  the  deflection  allowed 
is  0-3  inch. 

In  this  case,  ^=500,  /—  10,  ^=3100,  d=  6  and 
6  =  0-3  ;  and  by  substitution 

500  x  io3          _  50000  _ 
r  ~~~63x  0-3"  ~  2^088  ~ 


or   2\  inches  for  the  breadth. 

335.  —  Example  —  Value  of  d  in  a  Beam.  —  As  an  example 
under  formula  (126.},  find  the  depth  of  a  beam  of  maple  of 
average  quality,  which  is  5  inches  broad  and  20  feet  long, 
and  which  is  to  carry  3000  pounds  at  the  middle,  with  one 
inch  deflection. 

Here  we  have  F  —  $i$o,  W  —  3000,  /=2O,  £=5  and 
<J  =  i  ;  and  hence 

//  3OOO  X  203 

</=r  —       —  =  0.768 

5i$oxsx  i 
or  a  depth  of  Qf  inches. 


250 


RESISTANCE   TO    FLEXURE — RULES.         CHAP.  XVI. 


336.— Values  of  r,  n,  l>,  a  and  6  in  a  Lever.— The 
rules  for  the  quantities  in  a  semi-beam  or  lever  are  derived 
from  formula  (12 '!),  which  is 

i6/V 


and  are  as  follows : 


The  load, 


d  = 


Fbd3 


P  = 


Fbd36 
i6n3 


The  length,          n  =  V . 
The  breadth,        b  = 


d 


_  Vi6/V 


Fbd 


(123) 

(127) 

.     (128) 

(129) 


337.  —  Example—  Value  of  n  in  a  Lever.  —  As  an  example 
under  (127)  :  What  length  is  required  in  a  semi-beam  or 
lever  of  ash  of  average  quality,  3x7  inches  cross-section, 
and  carrying  200  pounds  at  the  free  end,  with  a  deflection 
of  half  an  inch  ? 

In  this  example,  P=  200,  ^=4000,  £  =  3,  d=  7  and 
6  =  o-  5  ;  and  we  have  • 


_  1/4000  x  3  x  73  x  p.  5  = 
16x200 

or  the  length  is  to  be   8   feet   J\  inches. 


_ 


338. — Example— Value  of  6  in  a  Lever. — Under  formula 
(128) :  What  is  the  proper  breadth  for  a  lever  of  hickory 
of  average  quality,  3  inches  deep,  projecting  4  feet  from 


DEFLECTION—  DIMENSIONS   OF   LEVER.  25  I 

the  wall  in  which  it  is  fixed,  carrying  a  load  of  200   pounds 
at  the  free  end,  and  having  a  deflection  of  one  inch  ? 

In  the  formula,    F  —  3850,   d—^,   6=  i,   p—  200    and 
n  =  4.     Substituting  these  values,  we  have 

> 
i6x  200  x43  _ 

"7-    ['97 


The  breadth  must  be  2  inches. 


339. — Example— Value  of  d  in  a  Lever. — What  must  be 
the  depth  of  a  bar  of  cherry  of  average  quality,  i^  inches 
broad,  projecting  3  feet  from  the  wall  in  which  it  is  im- 
bedded, and  carrying  at  its  end  a  load  of  100  pounds,  with 
a  deflection  of  f  of  an  inch  ? 

Here  P—  100,  n  =  3,.F=2%$o,  £=1-5  and  d  =  o-75; 
and  formula  (129.)  becomes 


d  =  V'     l6xIOQX33      =  2  • 

2850  x  i -5  x 0-75 


The  depth  required  is   2f   inches. 


340. — Deflection  —  Uniformly  Distributed  Load  on  a 
Beam. — The  cases  hitherto  considered  in  this  chapter  have 
all  had  the  load  concentrated  either  at  the  middle  of  a  beam 
or  at  the  end  of  a  lever.  When  the  weight  is  distributed 
equably  over  the  length  of  the  beam  or  lever,  the  deflection 
is  less  than  when  the  same  weight  is  so  concentrated. 

In  comparing  the  values  of  the  deflecting  energies  pro- 
ducing equal  deflections  in  the  two  cases,  we  have  [formula 
(511),  p.  477,  of  "  Mechanics  of  Engineering  and  Architec- 
ture," by  Prof.  Moseley,  Am.  ed.  by  Prof.  Mahan,  1856, 


2$2  RESISTANCE   TO   FLEXURE— RULES.         CHAP.  XVI. 

and  changing  the  symbols  to  agree  with  ours],  for  a  beam 
loaded  at  the  middle, 

WL3 

o  — 


and  [formula  (530.},  p.   484,   same    work],  for  a  beam  uni- 
formly loaded, 

UD 


Comparing  these  two  equal  values  of  eJ,    we  have 

WL3          UU 


Dr' 


or,  with  equal  deflections,  the  weight  at  the  middle  of  the 
beam  is  equal  to  f  of  the  uniformly  distributed  load. 
Thus,  100  pounds  uniformly  distributed  over  the  length 
of  a  beam  will  deflect  it  to  the  same  extent  that  62^  pounds 
would  were  it  concentrated  at  the  middle  of  the  length. 

Then,  since     U    represents  a  uniformly  distributed  load, 
%  U   will  equal  the     W  of  formula  (120.  \  which  formula  is 

Wl3 


Substituting  the  value  of   W,   as  above,  and  transposing,  we 
have 


for  the  relation  of  the  elements  in  the  deflection  jof  a  beam 
by  a  uniformly  distributed  load. 


DIMENSIONS   OF   BEAM — LOAD    DISTRIBUTED.  253 

341. — Value§  of  U,  19  b,  d  and  6  in  a  Beam. — By  in- 
versions of  formula  (130.)  we  have  the  following  rules — 
namely  : 

The  weight,        U  =  ^^  (131.) 


The  length,  /  = 

The  breadth,  b  =  ^  (133.) 

The  depth,  d  =  *^j^  (134.) 

The  deflection,  6  =  L^  (135) 


34-2.  —  Example—  Value  of    U,   the  Weight;  in  a  Beam.  — 

In  a  spruce  beam  of  average  quality,  20  feet  long  between 
bearings,  4  inches  broad  and  12  inches  deep  :  What  weight 
uniformly  distributed  over  the  beam  will  deflect  it  2  inches  ? 
In  this  example,  F  =  3500,  b  =  4,  d  =  12,  6  =  2  and 
/=  20;  and  by  formula  (131.) 


20 


or  the  weight  required  is    9677   pounds. 


343. — Example— Value  of   £,    the  Length,  in  a  Beam.— 

In  a  3  x  10  white  pine  beam  of  average  quality  :  What  is  the 
proper  length  to  carry  6000  pounds  uniformly  distributed, 
with  a  deflection  of  2  inches? 


254  RESISTANCE   TO   FLEXURE— RULES.         CHAP.  XVI. 

Here   F  =  2900,   b  —  3,   d  =  io,   6  —  2   and    U  =  6000 ; 
and  by  the  substitution  of  these  in  formula  (132.} 


A/2QOO 

=Y  -    — 


=16-68 


_ 

x  6000 
or  the  required  length  is    16   feet   8   inches. 


344. — Example— Value  of   &,    the  Breadth,  in  a  Beam. — 

Given  a  beam  of  average  quality  of  Georgia  pine,  20  feet 
long  and  10  inches  deep.  If  this  beam  carry  a  uniformly 
distributed  load  of  8000  pounds,  with  a  deflection  of  if 
inches,  what  must  be  the  breadth  ? 

We  have,  as  values  of  the  known  elements,  U  =  8000, 
1—20,  F=  5900,  d—  10  and  6=1.75;  and  formula  (133.) 
gives  us 

5  x  8000  x  2o3 
=  8  x  5900  x  io3  x  i~7s  =3-874 

The  breadth  must  be    3  j-   inches. 


34-5.  —  Example  —  Value  of   d,    Hie   Depth,  in   a  Beam.  — 

A  girder  of  average  oak,  8  inches  broad,  and  io  feet  long 
between  bearings,  is  required  to  carry  10,000  pounds  uni- 
formly distributed  over  its  length,  with  a  deflection  not  to 
exceed  -^  of  an  inch.  What  must  be  its  depth  ? 

The  elements  of  this  case  are  U  =.  10000,  /  =  io, 
F=  3100,  b  =  8  and  d  =  0-3.  Applying  formula  (13  4-)  we 
find 


or  we  must  make  the  depth   9^   inches. 


DEFLECTION — LOAD   DISTRIBUTED. 


255 


346.— Example— Value  of  (5,   the  Deflection,  in  a  Beam.— 

We  have  a  3  x  6  inch  beam  of  hemlock  of  average  quality, 
10  feet  long.  What  amount  of  deflection  would  be  produced 
by  3000  pounds  uniformly  distributed  over  its  length  ? 
£7=3000,  /— 10,  F  =  2800,  b  —  3  and  d  —  6;  and  the  for- 
mula applicable,  (13u.\  becomes 

5  x  3000  x  io3 


8  x  2800  x  3  x  63 " 
or  a  resulting  deflection  of    i    inch. 


347.— Deflection— Uniformly  Distributed  Load  on  a 
Lever. — For  a  load  at  the  free  end  of  a  lever  [Moseley's  Me- 
chanics (cited  in  Art.  340),  formula  (509. \  p.  476,  changing 
the  symbols]  we  have 

6  = 


and  [page  482,  same  work,  formula  (525.)~\  for  a  lever  with  a 
uniformly  distributed  load,  we  have 


6  = 

Comparing  these  equal  values  of   6   we  have 

PN3        UN3 

'-  TEI  Dr' 

u 

±  8        « 


or,  the  deflection  by  a  uniformly  distributed  load  is  equal  to 
that  which  would  be  produced  by  f  of  that  load  if  suspended 
from  the  end  of  the  lever. 


256  RESISTANCE  TO   FLEXURE — RULES.        CHAP.  XVI. 

348. — Values  of  C7,  n,  b,  d  and  fi  in  a  Lever. — In  for- 
mula (123.),  which  is  P  —  — g-y-,  we  have  the  relations  exist- 
ing between  the  elements  involved  in  the  case  of  a  lever 
under  strain. 

If  the  weight  uniformly  distributed  over  the  length  of  the 
lever  be  represented  by  £/,  then  P  =  -f  U  and  formula  (123.) 
becomes 


and  from  this  we  have  the  following: 

The  weight,  V  =  ~r  (136.) 


The  length,  n  =  \  -^r  (137^ 

The  breadth,  b  =  -^/^  (138.) 

The  depth,  d  = 


The  deflection,        d  =  -^s  (140.) 


349.— Example— Value  of   U,    the  Weight,  in  a  Lever.— 

In  a  Georgia  pine  lever  of  average  quality,  6  inches  broad 
and  10  inches  deep,  and  projecting  10  feet  from  the  wall 
in  which  it  is  imbedded  :  What  weight  uniformly  distributed 
over  the  lever  will  deflect  it  2  inches? 

In   this   example,    F=  5900,    b  =  6,    d=io,    6  =  2    and 
n  =  10 ;  and  by  formula  (136. \ 

TT      5ooox6x  io3x  2 

U  =  —  -=  11800 

OX   IO 


DIMENSIONS  OF  LEVER— LOAD   DISTRIBUTED.  257 

or  the  uniformly  distributed  weight  required  is  11,800 
pounds. 

Three  eighths  of  this  weight,  or  4425  pounds,  concen- 
trated at  the  free  end  of  the  lever,  will  deflect  it  the  same 
amount,  viz. :  2  inches. 

350.— Example— Value  of   n9    the  Length,  in  a   Lever.— 

In  a  lever  of  the  same  description  as  in  the  last  article,  except 
as  to  length  and  load  :  What  is  the  proper  length  to  carry 
8000  pounds  uniformly  distributed,  with  a  deflection  of  2 
inches  ? 

Here  we  have  ^=5900,  #  —  6,  d=  10,  6=2  and 
U=  8000 ;  and  by  the  substitution  of  these  in  formula  (137.) 


SQOO  x  6  x  io3  x  2         3.  -- 

=  n-383 


or  the  required  length  is   1  1    feet  4^  inches. 

351.—  Example—  Value  of   6,    the  Breadth,  in  a   Lever.— 

Given  a  lever  of  like  description  as  in  Art.  349,  except  as 
to  breadth  and  load.  If  this  lever  carry  a  uniformly  distrib- 
uted load  of  6000  pounds,  what  must  be  the  breadth? 

We  have,  as  values  of   the   known   elements,    U=  6000, 
n=io,    ^=5900,    d=  io    and   d=2\   and   formula  (138.) 

gives  us 

6  x  6000  x  io3 
*  =  —         —  i  -  =  3-051 

59OO  X  IO   X2 

The  breadth  must  be   3   inches. 

352.—  Example—  Value  of    d,   the  Depth,   in   a   Lever.— 

A  lever  of  like  description  as  in  Art.  349,  except  as  to  depth 
and  load,  is  required  to  carry  10,000  pounds  uniformly  dis- 
tributed over  its  length  :  What  must  be  its  depth  ? 


258  RESISTANCE   TO   FLEXURE — RULES.        CHAP.  XVI. 

The    elements   of   this    case    are      U  =  10000,      n  =  io, 
F  =  5900,   b  =  6   and   6  =  2.     We  apply  formula  (139.)  and 

find 

//6x  loooox  io3 

d  =  V  -  —? —  9  •  403 

5900  x  6  x  2 

or  we  must  make  the  depth  9^  inches. 

353.— Example— Value  of  d,   the  Deflection,  in  a  L-ever.— 

We  have  a  lever  of  like  description  as  that  in  Art.  349,  ex- 
cept as  to  load  and  deflection  :  What  amount  of  deflection 
would  be  produced  by  5000  pounds  uniformly  distributed 
over  its  length  ? 

£/r=  5000,   72—10,   7^=5900,   b  =  6  and  d=  io;   and  the 
formula  applicable,  (140),  becomes 

6  x  5000  x  ioa 

6=-  —  =  08475 

5900  x6x  io 

or  a  resulting  deflection  of  J  of  an  inch. 


QUESTIONS   FOR   PRACTICE. 


354. — Given  a  beam  loaded  at  middle :  What  are  the  rules 
by  which  to  find  the  weight,  length,  breadth,  depth  and  de- 
flection ? 

355. — Given  a  lever  loaded  at  the  free  end:  What  are  the 
rules  by  which  to  find  the  weight,  length,  breadth,  depth  and 
deflection  ? 

356. — In  a  beam  with  the  load  uniformly  distributed:  What 
are  the  rules  by  which  to  obtain  the  weight,  length,  breadth, 
depth  and  deflection? 

357. — In  a  lever  with  the  load  uniformly  distributed :  What 
are  the  rules  by  which  to  obtain  the  weight,  length,  breadth, 
depth  and  deflection  ? 


CHAPTER    XVII. 

RESISTANCE   TO   FLEXURE  —  FLOOR   BEAMS. 

ART.  358.  —  Stiffness    a    Requisite  in  Floor  Beams.  —  The 

rules  given  in  Chap.  VI.  for  the  dimensions  of  floor  beams 
are  based  upon  the  ascertained  resistance  of  the  material  to 
rupture,  and  are  useful  in  all  cases  in  which  the  question  of 
absolute  strength  is  alone  to  be  considered.  For  warehouses 
and  those  buildings  in  which  strength  is  principally  required, 
the  rules  referred  to  are  safe  and  proper  ;  but  for  buildings  of 
good  character,  in  which  the  apartments  are  finished  with 
plastering,  the  floor  timbers  are  required  to  possess  stiffness 
as  well  as  strength  ;  for  it  is  desirable  that  the  deflection  of 
the  beams  shall  not  be  readily  noticed,  nor  be  injurious  to 
the  plastering. 

359.  —  General  Rule  for  Floor  Beams.  —  The  relations  of 
the  several  elements  in  the  question  of  stiffness,  in  beams  uni- 
formly loaded  throughout  their  entire  length,  are  found  in 
formula  (130.), 

Fbd't 


The  load  upon  the  floor  beam  is  here  represented  by  U, 
and  its  value  is  U  =  cfl  (see  Art.  92)  ;  in  which  c  is 
the  distance  apart  between  the  centres  of  the  floor  beams, 
/  is  the  number  of  pounds  weight  upon  each  square  foot  of 
the  floor,  and  /  is  the  length  of  the  beam  ;  c  and  /  both 


GENERAL  RULE  FOR  FLOOR  BEAMS.         261 

being  in  feet.     If  for    U    we  substitute  this  value,  and  for   6 
put   rl  (see  Arts.  313  and  314),  we  have 

=  Fbd3r  (141 .) 


360.  —  The  Rule  modified.  —  For  the  floors  of  dwellings 
and  assembly  rooms,  ft  the  load  per  foot,  may  be  taken  (see 
Art.  115)  at  70  pounds  for  the  loading  and  20  pounds  for 
the  weight  of  the  materials,  or  90  pounds  in  all;  and  r,  the 
rate  of  deflection  per  foot  of  the  length,  at  0-03  (see  Art. 
314).  Formula  (141-)  thus  modified  becomes 

90  x  %cl3  = 

=  FM. 


8x0-03 

=  Fbd3 

cl3  = 


p 
This  coefficient,       ~7>    taking    F    at   its  average  value 


for  six  of  the  woods  in  common  use,  reduces  to 

firf  =3-15  for  Georgia  pine, 
=  2-69        "    locust, 
=  1-65        "   oak, 

f|4f  =1-87       "   spruce, 

=  1-55        "   white  pine, 
=  1-49        "   hemlock. 


361. — Rule  for   l>Aveliin^  and  Assembly   Rooms. — For 

p 
the  coefficient  in  (14&-),         ~^    putting  the  symbol     i,     we 


262  RESISTANCE  TO  FLEXURE — FLOOR  BEAMS.     CHAP.  XVII. 

have  this  simple  rule  for  problems  involving  the  dimensions 
of  floor  beams  in  dwellings  and  assembly  rooms,  namely, 

cl3  =  ibd3  (143.) 

and   we  have  the  value  of    i  for  average  qualities  of  six  of 
the  more  common  woods,  as  taken  in  Art.  360,  as  follows : 

For  Georgia  pine,  i  =  3-15 

"     locust,  i  =  2-69 

"     oak,  i  —  1-65 

"     spruce,  i  —  1-87 

"     white  pine,  i=  1-55 

"     hemlock,  i=  1-49 

362. — Rules  giving  the  Values  of  c,  J,  6  and  d. — Tak- 
ing formula  (14$ •)  we  derive  by  inversions  the  following 
rules,  namely : 

The  distance  from  centres,     c  =  -j-r  (144-) 


The  length,  /  =  \  -  (145.) 

C 


The  breadth,  b  =  j^3 

The  depth,  d  =  V~  (147.) 

363. — Example— Distance  from  Centres. — At  what  dis- 
tance from  centres  should  3x12  inch  Georgia  pine  beams 
of  average  quality,  24  feet  long,  be  placed  in  a  dwelling- 
house  floor? 

Here  we  have  2':=  3 -15,  £  =  3,  d  =  12  and  /=  24; 
and  by  formula  (144-) 

3-15  x  3  x  12*  _ 

or  the  distance  c  should  be  about    14^  inches. 


EXAMPLES    OF   DIMENSIONS.  263 

364. — Example— Length. — Of  what  length  may  average 
quality  white  pine  beams  3  x  10  inches  square  be  used,  when 
placed  16  inches  from  centres  ? 

In  this  case  2  =  1.55,  £  =  3,  d=io  and  c=  i^;  and 
formula  (14&*)  gives 


/ — 

-    ^3487-5  =  I5-I65 


or  these  beams  may  be  used   1 5    feet   2  inches  long  between 
bearings. 

365. — Example— Breadth. — In  floor  beams  20  feet  long 
and  12  inches  deep,  of  oak  of  average  quality,  placed  one 
foot  from  centres  :  What  should  be  the  breadth  ? 

Here,  c  —  i,  I—  20,  d  =  12  and  2=1-65.  With  for- 
mula (146. \  therefore,  we  have 


1-65  x  I23 
The  breadth  should  be  nearly   2-J ,   or  say    3    inches. 

366. — Example— Depth. — What  should  be  the  depth  of 
spruce  beams  of  average  quality  when  3  inches  broad  and 
20  feet  long,  and  placed  20  inches  from  centres?  The  sym- 
bols in  this  case  are  <r=if,  /=2O,  £=3,  and  i  =  1-87; 
and  by  formula  (14? •)  we  find 

W^ 

1-87x3 

or  the  depth  required  is  13!  inches.  Beams  3x13  could  be 
used,  provided  the  distances  apart  from  centres  were  cor- 
respondingly decreased.  The  new  distance  would  be  (form. 
144-)  !&2  instead  of  20  inches. 


264  RESISTANCE  TO  FLEXURE  — FLOOR  BEAMS.     CHAP.  XVII. 

367. — Floor  Beams  for  Store§. — The  several  values  of 
/  for  dwellings  and  assembly  rooms,  as  given  in  Art.  361, 
will  be  appropriate  also  for  stores  for  light  goods,  because 
timbers  apportioned  by  the  rules  having  these  values  of  t, 
will  bear  a  load  of  200  pounds  per  superficial  foot  before 
their  deflection  will  reach  the  limit  of  elasticity. 

For  first-class  stores — those  intended  for  wholesale  busi- 
ness, as  that  of  dry-goods — the  values  of  i,  as  above  given, 
are  too  large.  The  proper  values  for  this  constant  may  be 
derived  as  below. 


368. — Floor  Beams  of  First-elass  Stores. — The  load  upon 
the  floors  of  first-class  stores  may  be  taken  at  250  pounds  per 
superficial  foot,  and  the  deflection  at  0-04  of  an  inch  per  foot 
lineal  (see  Arts.  313  and  3I4-).  Beams  proportioned  by  these 
requirements  will  bear  a  load  of  about  3  x  250  =  750  pounds 
per  foot  before  the  deflection  will  reach  the  limit  of  elasticity. 
With  250  as  the  loading,  and,  say  25  pounds  (Art.  99) 
for  the  weight  of  the  materials  of  construction,  we  have 

7-275. 

Formula  (141  •)>  modified  in  accordance  herewith,  putting 
r  =  0-04,  becomes 

5  x  2'j^cl3  =  8  x  o-o<\Fbds 


369. — Rule  for  Beams  of  First-elass  Stores. — Reducing 

zp 

the   above   constant,   — ,  for   six   of  the   more   common 

4296^ 

woods  of  average  quality,  and  putting  the  symbol  k  for  the 
results,  we  have  for 


BEAMS   FOR   FIRST-CLASS    STORES.  265 

Georgia  pine,  k  =  1-37 

Locust,  k  —  i- 1 8 

Oak,  k  —  0-72 

Spruce,  £  =  0-81 

White  pine,  k  =  0-67 

Hemlock,  £  =  0-65 

With  this  symbol    k,   the  rule  for  floor  beams  of  first-class 
stores  is  reduced  to  this  simple  form, 

d3  =  kbd5  (149) 

370. — Values  of  c,  I,  b  and  d. — By  proper  inversions, 
we  obtain  from  formula  (149.),  rules  for  the  several  values 
required,  thus  : 

The  distance  from  centres,  c  =  —^-  (150.) 

The  length,  /  =  j/^!  (151.) 

The  breadth,  b  =  |~  (152) 

The  depth,  d  =  V-^  (153) 

371. — Example— Distance  from  Centres. — In  a  first-class 
store  :  How  far  from  centres  should  floor  beams  of  Georgia 
pine  of  an  average  quality  be  placed,  when  said  beams  are 
4  x  12,  and  20  feet  long  between  bearings? 

In  this  example,  we  have  k  =  i  -37,  b  =  4,  d  =  12  and 
/  —  20.  Then  by  formula  (150.) 


or  the  distance  from  centres  is   1-184  feet>  equal  to  about 
inches. 


266          .RESISTANCE  TO  FLEXURE — FLOOR  BEAMS.     CHAP.  XVIL 

372. — Example— Length. — At   what    length  may     4x10 

inch  beams  of  average  oak  be  used  in  the  floors  of  a  first- 
class  store,  when  placed  12  inches  from  centres?  Here  we 
have  £  =  0-72,  b  =  4,  ^=10  and  c—\\  and  by  formula 
(151) 


j       4/0-72  Xx  io 
/= 


-       -  =  14-23 
or  the  length  should  be    14   feet   3    inches. 

373.  —  Example—  Breadth.  —  The  floor  beams  in  a  first- 
class  store  are  to  be  20  feet  long  and  14  inches  deep,  of  white 
pine  of  average  quality.  When  placed  12  inches  from  cen- 
tres, what  should  be  their  breadth  ?  Taking  formula  (152.}  we 
have,  as  values  of  the  symbols,  c=i,  I  —  20,  k  =  o-6j  and 
^=1;  and 


TU  =  4-35 


0-67  x  14 
The  breadth  should  be   4^   inches. 

374. — Example— Depth. — What  should  be  the  depth,  in  a 
first-class  store,  of  spruce  beams,  of  average  quality,  4  inches 
thick  and  16  feet  long,  and  placed  14  inches  from  centres? 

In  this  case,  we  have  c  —  i|,  /  =  16,  k  =  0-81  and 
b  =  4.  Therefore,  by  formula  (153*) 


d  = 


0-81x4 

or  a  depth  of   I  if  inches. 

375. — Headers  and  Trimmers. — In  Chap.  VII.,  in  Arts. 
143  to  158,  rules  for  headers  and  trimmers,  based  upon  the 
resistance  of  the  material  to  rupture,  are  given.  These  rules 


STRENGTH   AND    STIFFNESS   COMPARED.  267 

contain  the  symbol  a,  which  represents  the  number  of  times 
the  weight  to  be  carried  is  contained  in  the  breaking  weight. 
The  value  of  this  symbol  may  be  assigned  at  any  quantity 
not  less  than  that  which  is  given  for  it  in  Table  XX.,  and, 
when  made  so  great  that  the  deflection  shall  not  exceed  0-03 
of  an  inch  per  foot  of  the  length,  the  rules  referred  to  will  be 
proper  for  use  for  headers  and  trimmers  for  the  floors  of 
dwellings  and  assembly  rooms. 


376. — Strength    and    Stiffness— Relation    of  Formulas.— 

The  value  of   a,   the  symbol  for  safety,  may  be  determined 
from  the  following  considerations : 
Taking  formula  (113.),  which  is 


"  bd'd 

and  substituting    G  for    W  we  have 

Gl3  =  Fbd3S 

A  comparison  of  the  constants  for   rupture    (B)    and  for 
elasticity   (F)   shows  that 

p 

B  :  F : :  I  :  m  =  -^ 

or  Bm  =  F 

and  putting    rl    equal  to    d    we  have,  by  substitution, 

Gl3  =  Bmbd'rl 
Gl2  =  Bmbd'r 


dmr 


268  RESISTANCE  TO  FLEXURE— FLOOR  BEAMS.     CHAP.  XVII, 

We  have  by  formula  (10.),  Art.  36, 

Wl 

*w 

or  Wl  =  Bbd2 

Comparing  this  value  of    Bbd*  with  that  above,  we  have 

Gla 


Wl  = 


dmr 


In  this  formula,  G  is  the  weight  which  may  be  carried  by 
the  beam,  with  a  deflection  per  foot  of  the  length  equal  to  r; 
and  W  is  the  breaking  weight.  Putting  these  symbols  in  a 
proportion,  we  have 

W 
G-.  W::  i  :a=^r 

•  G 

or  Ga  =  W 

Substitute  for    W  this  value  of  it,  and  we  obtain 

r  j       Gr 
Gal  =  -j — 

dmr 
I 


-j —      — ~ 
dmr         F 
d-Br 


377. — Strength  and  Stiffki ess— Value  of   <i,  in  Terms  of  B 

and    F.  —The  values  of   B   and   F  (form.  154-}  are  found  in 

Table  XX.,  and    r  =  0-03.     The  ratio    -7-   (  /    in  feet  and     d 

in  inches)  cannot  be  exactly  determined  until  the  length  and 
depth  have   been    established.     An   approximation    may  be 


MEASURE   OF  THE   SYMBOL   FOR   SAFETY.  269 

assumed,  however,  for  a  preliminary  calculation,  and  then, 
if  found  to  err  materially,  it  may  be  taken  more  nearly  cor- 
rect in  a  final  calculation.  In  all  ordinary  cases,  the  ratio 

-T-  will  be  found  nearly  equal  to  |£  =  i  -  7.  Taking  this  value 
in  formula  (IS  4-)  we  have 


378.  —  Example.  —  Let  us  apply  this  in  the  use  of  formula 
.),  namely  : 


Wai  =  Bbd* 


What  weight  may  be  carried  at  the  middle  of  a  Georgia 
pine  beam  of  average  quality,  3  x  10  inches  x  17  feet,  so  as 
to  deflect  it  no  more  than  would  be  proper  for  the  floors  of 
a  dwelling? 


Here   £  =  3,  d=  10,   a—  -  ^     ,    ^=5900  and    I =  17  ; 

therefore 

B  x  3  x  io2  _  5900  x  3  x  ioa 

yy   -— - 


jjr    17 
i  770000 


379.  —  Te§t  of  the  Rule.  —  To   test   the    accuracy  of  the 
result  just   found,   the   same    problem    may   be   solved    by 

formula  (113.), 

W>_ 

F''~~bd3d 

from  which  we  have,  when    <?  =rl,    and  substituting    G   for 

W> 

Gl*  =  Fbdsr 

_      Fbd'r 
and  6-  —  —  j-2  — 


2/0  RESISTANCE  TO  FLEXURE— FLOOR  BEAMS.     CHAP.  XVII. 

In  this  expression,  in  the  above  example,   F  =  5900,   b  =  3, 

d=  10,    1=17   and   r  =  0-03  ;    and  hence 

t 

5900  x  3  x  10s  x  0-03        531000 

""  ~~    =l837'4 


380.  —  Rules   for    Strength    and    Stiffne§§   Resolvable.  — 

The  result  in  the  last  article  is  the  same  as  the  one  before 
found,  and  indeed  could  not  be  otherwise,  since  the  one 
formula  is  derived  directly  from  the  other,  and  is  readily 
resolvable  into  it  ;  for  if,  in  formula  (21.), 

Wai  =  Bbd2 

we  substitute  for  a  its  equivalent  as  in  formula  (154*),  we 
have 


Fdr 
Wl*  =  Fbd'r 

so  that  instead  of  computing  the  value  of  a  for  use  in  any 
particular  case  by  formula  (155.),  we  may  introduce  into  the 
rule  its  value  as  given  by  (154),  and  reduce  to  the  lowest 
terms,  as  in  the  next  article. 


381.  —  Rule  for  the  Breadth  of  a  Header.  —  A  rule  for  a 
header  is  given  in  formula  (27  .\  Art.  145.  Substituting  for 
a  its  value  as  in  (154°),  we  have,  taking  ^U  instead  of  J£/,  Art. 
340, 


In  this  expression,  /  and  g  are  the  same,  both  represent- 
ing the  length  of  the  header,  and  the   (d—\)  is  put  for  the 


HEADERS  FOR  DWELLINGS,  271 

effective  depth,  and  is  equal  to  the  d  of  the  first  member; 
therefore,  reducing,  we  have 


which  is  a  rule  for  the  breadth  of  a  header,  based  upon  the 
resistance  to  flexure. 


382.  —  Example    of  a    Header    for    a    Dwelling.  —  In   a 

dwelling  having  spruce  floor  beams  of  an  average  quality, 
10  inches  deep  :  What  would  be  the  required  breadth  of  a 
header  of  the  same  material,  10  feet  long,  carrying  tail 
beams  12  feet  long? 

The  values  of  the  symbols  are,  f=  90  (Art.  115),  n  =  12, 
£•=10,   ^=3500   (Table   XX.),   r  =  0.03    and   d=io\    and 

,  5  xgox  12  x  io5 

b  =  —?—  —9  =  4-409 

16x3500x0-03  XQ 

or  the  required  breadth  is  4|  inches  full. 


383.  —  Example  of  a  Header  in  a  Fir§t-clas§  Store.  —  In  a 

first-class  store,  where  the  beams  are  14  inches  deep,  what 
is  the  required  breadth  of  a  header  of  Georgia  pine  of  aver- 
age quality,  16  feet  long,  and  carrying  tail  beams  17  feet 
long  ? 

Here    /=  275,     r  =  0-04     (Art.  368),    n  =  17,    g  —  16, 
F  =  5900  (Table  XX.)  and  </=  14;  and  by  formula  (156.  \ 

,  5x275x17x16" 

b    =   ~~  -  —  -  '-  -  3=     II'54I 

16  x  5900  xo-O4x  i33 
The  breadth  should  be  1  1|  inches  full. 


272  RESISTANCE  TO  FLEXURE  —  -FLOOR  BEAMS.     CHAP.  XVII. 

384.  —  Carriage  Beam  with  One  Header.  —  (See  Art. 
389.)  In  Art.  ISO  a  rule  (form.  29.)  is  given  for  this  case, 
based  upon  the  resistance  of  the  material  to  rupture.  As 
with  a  header  (Art.  381),  so  here,  the  rule  given  may  be 
resolved  into  one  depending  upon  the  resistance  to 
deflection. 

Taking  formula  (29.),  and  for  a  substituting  its  value  as 
per  formula  (154-),  we  have,  taking  ££/  instead  of  ££/,  Art.  340, 


)  =  Fbd*r  (157.) 

which  is  the  required  rule. 

385.  —  Carriage  Beam  with  one  Header,  for  I>welling§. 

In  this  rule,  putting  f=  90  and  r  =  0-03,  we  obtain 

3000  (bcl*+gn*m)  =  Fbd* 


which    is   a    rule    for    carriage    beams   with    one    header,    in 
dwellings  and  assembly  rooms.     (See  Art.  389.) 

386.  —  Example.  —  What  should  be  the  breadth,  in  a 
dwelling,  of  a  carriage  beam  of  average  quality  white  pine, 
20  feet  long  by  12  inches  deep,  and  carrying  a  header  16 
feet  long  at  a  point  5  feet  from  one  end  ?  The  floor  beams 
among  which  this  carriage  beam  is  placed  are  set  at  16 
inches  from  centres. 

Here  c  —  i^,  /  —  20,  g  =  16,  n  =  15,  m  —  5,  F  =  2900 
and  d  =  12  ;  and  by  formula  (158.) 


b  =  ^ 

2900  X  I28 
The  breadth  should  be  12}  inches. 


CARRIAGE   BEAMS   WITH   ONE   HEADER.  273 

387.  —  Carriage  Beam  with  One  Header,  for  First-class 
Stores.  —  If  in  formula  (157.)  we  take  the  value  of  /  equal 
to  275,  and  of  r  equal  to  0-04,  we  shall  then  have 

6875  (ficl*+gn%m)  =  Fbd* 


which  is  the  required  rule  (see  Art.  389). 


388.  —  ExampDc.  —  Of  what  breadth,  in  a  first-class  store, 
should  be  a  Georgia  pine  carriage  beam  of  average  quality, 
25  feet  long,  and  carrying  at  6  feet  from  one  end  a  header 
16  feet  long;  the  floor  beams  being  15  inches  deep,  and 
placed  15  inches  from  centres? 

Here  c  =  ij,  /  =  25,  g  —  16,  n  —  19,  m  =  6,  d  =  15 
and  F  =  5900  ;  and  formula  (159.)  becomes 


6875  [(A  x  i}  x  253)  +  (16  x  iQ2  x  6)] 
£  —  —  /  J  LMT  -  2  -  /j  _ 

5900  X  i$3 
or  the  breadth  required  is  14^  inches. 


389.  —  Carriage  Beam  with  One  Header,  for  Dwellings- 
More  Precise  Rule.  —  The  rules  above  given  (157.,  158, 
and  159.)  are  not  strictly  correct  :  they  give  a  slight  excess 
of  material  (see  Art.  241). 

The  rule  shown  in  formula  (86.),  taking  f  £/",  Art.  340, 


is  accurate*  and  should  be  the  one  employed  in  special  cases 

*  Except   when    h   is   less   than   n   {Art.    240).      In   this    case   the   result  is 
slightly  in  excess,  but  so  slightly  that  the  difference  is  unimportant. 


274  RESISTANCE  TO  FLEXURE—  FLOOR  BEAMS.     CHAP.  XVII. 

in   which    a    costly  material    is    used.      Substituting  for  a  in 
this  formula,  its  value,  as  in  formula  (154>),  we  have 

Bl  mn  . 


A'  +  f  U)  =  Fbd*r  (160.) 

in  which  A'  is  the  concentrated  load,  and  U  the  uniform- 
ly distributed  load.  Formula  (160.)  may  be  modified,  in  the 
case  of  a  carriage  beam,  by  using  for  these  symbols  their 
values,  thus: 

From  Arts.  92    and    150,    A'  =  ±fng,    and    U=$cfl,    and 

hence 

fmn  (ng  +  %cl)  =  Fbd*r  (161.) 

which  is  a  more  precise  general  rule  for  a  carriage  beam 
carrying  one  header. 

If,    now,   we    put  f  equal    to   90,    and    r   equal    to   0-03, 

we  shall  have 

yxx>mn  (ng  4-  \cl)  —  Fbd* 


,       yxxmn  (ng  +  frZ) 
Fd* 

which  is  a  more  precise   rule   for   carriage   beams  with   one 
header,  in  floors  of  dwellings  and  assembly  rooms. 


390.  —  Example.  —  Taking  the  example  given  in  Art.  386, 
we  have  m  =  5,  «=I5,  £•=  16,  r  =  ij,  /  =  20,  ^=2900 
and  d  =  12  ;  and,  in  formula  (162.) 


=  I2> 

2900  X  I23 

showing  that  by  this,  the  more  exact  rule,  the  breadth 
should  be  \2\  inches,  while  by  the  former  rule  it  was  deter- 
mined to  be  I2j  inches. 


PRECISE   RULE   FOR   CARRIAGE   BEAMS.  275 

391. — Carriage  Beam  witli  one  Header,  for  First-class 
Stores— More  Precise  Rule. — Modifying  formula  (161.),  by 
putting  275  for  /",  and  0-04  for  r,  we  have 

6875***  (rig  +  Id)  =  Fbd* 


Fd"  (1631 

which  is  the  more  precise  rule  required. 

392. — Example. — Applying  this  rule  to  the  example 
given  in  Art.  388,  we  find,  m  —  6,  n  —  19,  g—  16,  c  —  ij, 
I  —  25,  F—  5900  and  d  =  15  ;  and  hence 

^6875  x6x  19(19  x  i6  +  f  x  ijx  25)  _ 
5900 x  I53 

giving  the  breadth,  by  this  more  precise  rule,  at  13^  inches. 
This  is  nearly  half  an  inch  less  than  by  the  former  rule, 
which  gave  for  the  breadth,  14-073,  or  14^  inches  nearly. 

393. — Carriage  Beam  with  Two  Headers  and  Two  Sets 
of  Tail  Beams,  for  Dwelling*,  etc. — Formula  (32.)  in  Art.  155 
gives  the  relations  of  the  symbols  referring  to  a  case  in 
which  a  carriage  beam  has  to  carry  two  headers,  with  two 
sets  of  tail  beams.  From  this  formula  we  have,  taking  f  U, 
Art.  340, 

b  —  TrWVfi 


If  in   this   equation   the  value  of  a,  as  in  formula  (154-), 
be  substituted,  there  results 


which  is  a  general  rule  for  these  cases. 


2/6  RESISTANCE  TO  FLEXURE — FLOOR  BEAMS.     CHAP.  XVII. 

Putting/ =90   and    r  —  o-o^,    we  have 

b  =  ~?  \_grn  (mn  +  s2)  +  TV/8]  (165.) 

which  is  a  rule  for  a  carriage  beam,  carrying  two  headers, 
with  two  sets  of  tail  beams,  in  the  floor  of  a  dwelling  or 
assembly  room.  (See  Arts.  402,  405,  415  and  417.) 

394. — Example. — Under  rule  (165.)  take  the  example 
given  in  Art.  156,  in  which  F  =  5900,  ^=14,  g  •=.  12, 
c  =  i-J-  and  /=25.  For  m  and  s  there  are  given  5  and 
15,  and  taking  m  as  the  larger,  m  =15,  n  =  10,  s  =  5 
and  r  =  20  ;  so  that  (165.)  becomes 


or  the  breadth  should  be  j\  inches. 

395. — Carriage  Beam  with  Two  Headers  and  Two  Sets 
of  Tail  Beams,  for  First-elass  Stores. — If,  in  formula  (164), 
f  be  put  at  275  and  r  at  0-04,  we  shall  have 

6875 

which  is  a  rule  for  a  carriage  beam  carrying  two  headers, 
with  two  sets  of  tail  beams,  in  a  first-class  store  (see  Arts. 
402,  407  and  4(7). 

396. — Example. — Referring  to  the  same  example  (Art. 
156)  we  have  F  =  5900,  d  —  14,  g=.  12,  m  —  15,  n  =  10, 
s  =  5,  c  =  i£  and  /  =  25  ;  and  the  formula  is 


b  =  5QOx5i43[12  x  I5  (I5  x  1Q  +  52)  +  TVx  H  x  253]  =  16-486 
or  the  breadth  should  be   i6£  inches. 


CARRIAGE   BEAM   WITH   TWO   HEADERS.  277 

397.  —  Carriage  Beam  with  Two  Headers  and  One  Set 
of  Tail  Beams.  —  Formula  ($4-),  in  Art.  157,  is  a  rule  for  a 
carriage  beam  with  two  headers,  carrying  but  one  set  of  tail 
beams.  Substituting,  in  this  formula,  for  a  its  value 

/?/ 
(form.  154.)  --  ,  we  have,  taking  §U,  Art.  340, 


from  which 


which  is  a  general  rule  for  a  carriage  beam  carrying  two 
headers,  with  but  one  set  of  tail  beams,  with  a  given  rate  of 
deflection.  (See  Arts.  402,  409,  411,  419  and  421.) 

398. — Carriage  Beam  with  .Two  Headers  and  One  Set 
of  Tail  Beams,  for  Dwellings. — If,  in  formula  (167.),  f  be 
put  at  90  and  r  at  0-03,  we  shall  have 


b  =  {Jgm  (n+s)  +  fcl*]  (168.) 


a  rule  for  a  carriage  beam  with  two  headers,  carrying  only 
one  set  of  tail  beams,  in  a  dwelling  or  assembly  room.  (See 
Arts.  402,  409,  411,  419  and  421.) 

399.  —  Example.  —  Let  it  be  required  to  find,  under  this 
rule,  the  breadth  of  a  carnage  beam  20  feet  long,  of  spruce 
of  average  quality  ;  said  beam  carrying  two  headers,  each 
12  feet  long,  with  tail  beams  11  feet  long  between  them, 
leaving  an  opening  4  feet  wide  on  one  side,  and  another 
5  feet  wide  on  the  other  side.  The  beams  among  which 
this  carriage  beam  is  placed  are  12  inches  deep  and  16 
inches  from  centres. 


278  RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII. 

For  the  symbols  we  have,  ^=3500,  d  =  12,  j—ii, 
g—\2,  c  =  1%  and  /  =  20.  Having  for  m  and  s  the 
values  4  and  5,  we  make  m  equal  to  the  larger  one,  and 
therefore  m  =  5,  n  —  15  and  s  —  4.  These  values  substi- 
tuted in  formula  (168.)  produce 

_  3000  [i  i  x  12  x  5  (i  5  +  4)  +  A  x  ij  x  2Q3] 

3500  XI23  -  7'8;4 

The  breadth  should  be,  say  7^  inches. 

400.—  Carriage  Beam  with  Two  Headers  and  One  Set 
of  Tail  Beams,  for  First-class  Stores.  —  If,  in  formula  (167.), 
we  put  275  for  /  and  0-04  for  r,  we  shall  have 

(169.) 


which  is  a  rule  for  carriage  beams  carrying  two  headers, 
with  one  set  of  tail  beams  between  them,  in  a  first-class  store. 
(See  Arts.  4-02,  409  and  413.) 

401.  _  Example.  —  What  should  be  the  breadth,  under 
this  rule,  of  a  carriage  beam  of  average  quality  Georgia 
pine,  25  feet  long,  with  two  headers  each  20  feet  long, 
carrying  tail  beams  10  feet  long  between  them  ?  The  tail 
beams  are  so  located  that  there  is  an  opening  10  feet  wide 
at  the  left-hand  end,  and  one  5  feet  wide  at  the  right-hand 
end.  The  tier  of  beams  is  15  inches  deep  and  placed  15 
inches  from  centres. 

Here  F=  5900,  d  =  i$,  j  =  10,  g  =  20,  c—i\  and 
1—2$.  For  the  values  of  m  and  s  we  have  10  and  5; 
and  10  being  the  larger  it  follows  that  m=  10,  n—  15 
and  s  =•  5  ;  and  by  formula  (169.), 


=  Ii;    lg 
5900  x  1  58 

or  the  breadth  should  be   15!  inches. 


MORE   PRECISE    RULES    FOR-  CARRIAGE   BEAMS.  279 

4-02.—  Carriage  Beam  with  Two  Headers  and  Two 
Sets  of  Tail  Beams—  More  Precise  Rules.  —  The  rules  for 
carriage  beams  given  in  Arts.  393  to  401  are  drawn  from 
formulas  which  arc  but  close  approximations  to  the  truth. 
The  resulting  dimensions  are  always  in  excess  slightly  of  the 
true  amounts,  and  the  rules  therefore  are  safe. 

The  rule  embodied  in  formula  (92.),  however,  is  deduced 
from  exact  premises,  and  its  results  are  precise. 

If  for  a  its  value  (form.  1£>4*)  b.e  substituted  in  formula 
(92.),  we  shall  have,  taking  f  C7,  Art.  340, 


(170.) 
and,  as  auxiliary  thereto, 


-~(rs  + 


When    h    is  equal   to  or  exceeds    n,    then    n    is   to  be 
substituted  for    h,   and  the  portion 


of  formula  (170.)  equals    a'  (see  Art.  248),  and  the  formula 
itself  reduces  to 


280          RESISTANCE  TO  FLEXURE—  FLOOR  BEAMS.     CHAP.  XVII. 
Substituting  for  a'  its  value  (form.  171.)  we  have 

b  =  - 


We  have  here,  in  formula  (170.),  a  general  rule,  and  in 
formula  (174-),  a  rule,  general  when  h  equals  or  exceeds 
n,  for  a  carriage  beam  carrying  two  headers,  with  two  sets 
of  tail  beams,  with  a  given  deflection. 

403.  —  Example—/*,  les§  than  n.  —  Let  it  be  shown,  under 
these  rules,  what  should  be  the  breadth  of  a  carriage  beam 
of  spruce  of  average  quality,  20  feet  long  and  12  inches 
deep,  carrying  two  headers  each  12  feet  long,  so  placed  as 
to  leave  an  opening  41  feet  wide  ;  said  opening  being  7^ 
feet  distant  from  one  wall  and  8  feet  from  the  other. 

The  floor  is  to  carry  100  pounds  per  superficial  foot, 
with  a  deflection  of  0-03  per  foot,  and  the  beams  are  placed 
15  inches  from  centres. 

Here  we  have  /=  100,  g=  12,  m  —  8,  !  =  20,  n  —  12, 
s=7h  r=\2^y  *=ii,  d'  =  l-(m+s)  =  20  —  (8  +  ;£)  = 
20—  I5i  —  4j,  ^=3500  and  d=  12. 

Preliminary  to  finding  the  value  of  h  we  have  to  deter- 
mine the  values  of  a'  and  b'  . 

By  formulas  (171.)  and  (172.) 

ICO  X  12  X  8 


a ' ''         4x20       (8xI2  +  7'53)       =18270 


ICO  X  12  X  7-5 


b'-~          4x20 ,    -("-S*  7-5  +  «•)=  17746-875 


a'  —  b' =      523-125 


CARRIAGE  BEAM — SPECIAL  RULES.  28 1 

From  these  and  formula  (173.}  we  have 


So    h  =  11-49,    an<3  since  it  is  less  than    n  (as    n    equals    12) 
is  therefore  to  be  retained ;  and  we  have  (form.  170.) 

-   fVx  i^x  100 x  n  -49x8- 51 +  17746 -875  + 

*    ^  O  *  O^  L— 


3500  X  I2J 

523.125  :~\ 

^y^X(!I.  49-7.  5)J   =9.714 

or  the  required  breadth  is  9f  inches. 


404. — Example—  h  greater  than  n. — What  should  be 
the  breadth  of  a  white  pine  carriage  beam  20  feet  long,  12 
inches  deep,  and  carrying  two  headers  10  feet  long — one 
located  at  9  feet  from  one  wall  and  the  other  at  6  feet  from 
the  other  wall ;  the  floor  to  carry  100  pounds  per  foot  super- 
ficial, with  a  deflection  of  0-03  of  an  inch  per  foot  lineal, 
and  the  beams  to  be  placed  15  inches  from  centres  ? 

Here  /=  100,  F  —  2900,  d  —  12,  r  =  0-03,  c  —  i-J, 
1=20  and  £-=10.  Comparing  m  and  s  we  have  m  =  9, 
n  =  1 1  and  s  =  6. 

Proceeding  as  in  the  last  article,  we  find  that  h  exceeds 
n,  therefore,  according  to  Art.  402,  we  have  formula  (174>) 
appropriate  to  this  case  ;  from  which 


*  =  2900x^x0.03  [(**  i*x  it  xao)  +  io(9x  u+6*)]  =  10-140 
or  the  breadth  should  be    loj   inches. 


282  RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII. 

405.  —  Carriage  Beam  with  Two  Headers  and  Two  Sets 
of  Tail    Beams,   for  Dwellings—  More   Precise  Rule.  —  If,  in 

formula  (17  4>),  f  =  90    and    r  =  0-03,    we  shall  have 


which  is  a  precise  rule  for  carriage  beams  carrying  two 
headers,  with  two  cets  of  tail  beams,  in  dwellings  and  assem- 
bly rooms.  (See  Arts.  393  and  402.) 

406.  —  Example.  —  An  example  under  this  rule  may  be 
had  in  that  given  in  Art.  404  ;  in  which  we  have  F=  2900, 
d=  12,  c  =  ij,  /  —  20,  ^=10,  m  =  9,  n  —  n  and  s  =  6. 
Then  by  formula  (175.} 


2a  Ki  x  r±  x  J  J  x  20)4-10(9x11  +62)]  =  9.126 
or  the  breadth  should  be  9^  inches. 

407,  —  Carriage  EScam  with  Two  Headers  and  Two  §et§ 
of  Tail  Beams,  for  First-class  Stores—  More  Precise  Rule.— 

If,  in  formula  (174.),   f—  275    and    r  =  0-04,    we  shall  have 


V         •—  J-,        J    Q 

Fd* 

which  is  a  precise  rule  for  carriage  beams  carrying  two 
headers,  with  two  sets  of  tail  beams,  in  first-class  stores. 
(See  Arts.  395  and  402.) 

408. — Example. — What  should  be  the  breadth,  under 
this  rule,  of  a  carriage  beam  of  Georgia  pine  of  average 
quality,  23  feet  long,  14  inches  deep,  carrying  two  headers 
each  17  feet  long,  with  tail  beams  on  one  side  7  feet  long, 
and  on  the  other  10  feet  long ;  the  beams  being  placed  14 
inches  from  centres  ? 


CARRIAGE    BEAMS   FOR    FIRST-CLASS    STORES.  283 

Here    ^=5900,    d  —  14,    c  =0i%>    /  =  23     and    g—\T. 
Taking  the  larger  of  the  two,    10    and    7,    for   ;«,    we  have 

m  =  10,    n  =  13,    and    s  =  7  ;    and  by  formula  (176.) 


14. 
the  breadth  should  be,  say     14!     inches. 


=  14-774 


409.  —  Carriage  Beam  with  Two  Headers  and  One  Set 
of  Tail  Beams—  More  Precise  Rule.  —  In  a  case  where  there 
are  two  openings  in  the  floor,  one  at  each  wall,  then  the  two 
headers  carry  but  one  set  of  tail  beams,  and  these  are  be- 
tween the  headers.  The  load  at  each  header  is  the  same  ; 
and  when  g  equals  the  length  of  header,  j  the  length  of 
tail  beams,  and  /  the  load  per  superficial  foot,  then  the 
load  at  each  end  of  each  header  is 

W=\fgj 

and  the  expression  for  the  load  at  one  point,  as  in  Art.  (53, 

wi  IVi'fz 

-j-(Wn+Vs\    becomes    --j--(*'+  f),    and  therefore  (A  rt.  243) 


(177.) 


and  fi'  =          (r  +  w)  (178.) 

4/ 

In  the  case  under  consideration,  these  two  expressions  are 
auxiliary  to  formula  (170.),  in  the  place  of  those  given  in  for- 
mulas (171.)  and  (172.),  and  with  h  equal  to,  or  exceeding 
n,  formula  (170.)  becomes 


284  RESISTANCE  TO  FLEXURE—  FLOOR  BEAMS.     CHAP.  XVII. 

Substituting  for   a'   its  vajue,  as  in  formula  (177.)t  we  have 


(179.) 


which  is  a  precise  rule  for  carriage  beams  carrying  two 
headers,  with  one  set  of  tail  beams,  and  with  a  given  rate  of 
deflection.  (See  Arts.  397,  398  and  402.) 


410. — Example. — What  should  be  the  breadth  of  a  car- 
riage beam  of  locust  of  average  quality,  16  feet  long  and  8 
inches  deep,  carrying  two  headers  of  8  feet  length,  with 
one  set  of  tail  beams  7  feet  long  between  them,  so  placed  as 
to  leave  an  opening  of  6  feet  width  at  one  wall,  and  another 
of  3  feet  at  the  other?  The  floor  beams  are  placed  15 
inches  from  centres,  and  are  to  carry  90  pounds  per 
superficial  foot,  with  a  deflection  of  0-04  of  an  inch  per 
foot  lineal. 

We  have  from  this  statement  f  =  90,  m  =  6,  ;/  =  10, 
/=  16,  r  =  13,  5=3,  c=  ij,  F=  5050,  d  =  8,  r7  =  0-04, 
g  =  8  and  j  =  7. 

To  test  the  value  of  h  we  have,  preliminary  thereto, 
formula  (177.),  which  gives 

oox  8  x  7x  6      

a  = 2- x  10  +  3  =  6142  •  5 

and,  formula  (178.), 

90x8x7x3      — - - 


V 

a'-V  =  1653.75 


CARRIAGE   BEAMS  FOR  DWELLINGS.  285 

Then,  by  formula  (173.), 


As    n  =  10,    h   exceeds   n.      We   must,    therefore,    substi- 
tute   n   for  h  ;     and  by  formula  (179.)  we  have 


or  the  breadth  should  be    5J,    say    5    inches. 


4(1. — Carriage  Beam  witli  Two  Header§  and  One  Set 
of  Tail  Beams,  for  Dwellings— lHore  Precise  Rule. — If,  in 

formula  (179.),  f  =  90   and    r'  =  0-03,   we  shall  have 

(180.) 

which  is  a  precise  rule  (in  cases  where  h  exceeds  n  )  for 
carriage  beams  carrying  two  headers,  with  one  set  of  tail 
beams,  in  a  dwelling  or  assembly  room.  (See  Arts.  398, 
402  and  409.) 


4(2. — Example. — What  should  be  the  breadth,  in  a 
dwelling,  of  a  carriage  beam  of  spruce  of  average  quality, 
1 8  feet  long  and  10  inches  deep,  carrying  two  headers  of 
12  feet  length,  with  a  set  of  tail  beams  between  them  7  feet 
long?  The  headers  are  placed  so  as  to  leave  an  opening  of 
8  feet  on  one  side  and  3  feet  on  the  other,  and  the  beams 
are  set  15  inches  from  centres. 

Here  /  =  90,  g  —  12,  j  =  7,  m  =  8,  n  —  10,  s  =  3, 
r  =  15,  /  =  18,  F  =  3500,  d  —  10,  r'  —  0-03  and  c  =  \\. 


286  RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII 

Preliminary  to  seeking  the  value  of   h    we  find,  by  for 
mulas  (177.)  and  (17  8.  \ 


,      90x12x7x8. 

"  ~- 


=  7245 


a'-b'=    3675 
Now,  by  formula  (173.), 


But  n=  10;  therefore  n  is  to  be  used  in  the  place  of  7z, 
and  formula  (180.)  is  the  proper  one  to  use  in  this  example. 
This  latter  formula  gives  us 

x8   ; 


10x18)  +  (12x7x10  +  3)]  =9-  417 


Thus    the    breadth    should    be    9!    inches  ;    or   the    beam   be 
of  x  10   inches. 


4-13.  —  Carriage  Beam  wittli  Two  Headers  ancl  One  Set 
of  Tail  Beams,  for  First-elass  Stores  —  More  IPrecfse  RuBc.  — 

If,  in  formula  (17  9.)  t   f—  275    and    r  —  0-04,    we  shall  have 


which  is  a  precise  rule,  when  h  exceeds  ?z,  for  a  carriage 
beam  carrying  two  headers,  with  one  set  of  tail  beams,  in  a 
first-class  store.  (See  Arts.  400  and  402.) 


CARRIAGE   BEAM   WITH   TWO    HEADERS.  287 

i 

4-14. — Example. — The  example  given  in  Art.  412  may  be 
used  to  exemplify  this  rule,  excepting  the  depth,  which  we 
will  put  at  14  inches  instead  of  10. 

Formulas  (180.)  and  (181.)  are  alike,  with  the  exception 
of  the  numerical  constant.  The  result  found  in  Art.  4-12, 
b  =  9-417,  multiplied  and  divided  to  correct  the  constant, 
will  give  the  result  required  in  this  case.  The  constant 
6875  is  to  take  the  place  of  3000,  and  the  depth  14  is  to 
replace  10.  With  these  changes,  we  have 

6875       TOGO 

b  —  9-417  x  — —  x  -     -  =  7-865 
3000       2744 

or  the  breadth  should  be    7-86;    say    7J   inches. 


415.  —  Carriage   Beam  with    Two    Headers,  Equicli§tuiit 
from  Centre,  and  Two  Sets  of  Tail  B  earn  §—  Precise  Rule.  — 

In  case  the  opening  in  the  floor  be  at  the  middle,  leaving 
tail  beams  of  equal  length  on  either  side,  then  the  moments 
of  the  two  concentrated  loads  upon  the  carriage  beam  are 
equal,  or  a'  =  b'  and,  in  formula  (170.), 


and  the  formula  itself  becomes 


in  which   b'   represents  the  combined  effect  of  the  two  loads, 
as  acting  at  the  location  of  either  of  them. 
This  effect  is  shown  (Art.  153)  to  be 


288  RESISTANCE  TO  FLEXURE—  FLOOR  BEAMS.     CHAP.  XVII. 

In  the  case  under  consideration,  W  —  V  and  m  =  s,  and 
therefore 

£'=  W~(n  +  m)  =   W-~   =   Wm 

Now,  W  represents  the  weight  concentrated  in  one  end  of 
one  of  the  headers.  The  load  on  a  header  is  %fgm,  and 
the  load  at  one  end  of  the  header  is  \fgrn  ;  therefore 

b'  = 
and  formula  (182.)  becomes 


*=Tfi 
By  formula  (178.) 


a'-b' 


and  since  in  this  case     a'  —  b'  =  o 


—  ^l—t  and 

and  therefore 


which  is  a  precise  rule  for  carriage  beams  carrying  two 
headers,  equidistant  from  the  centre,  with  two  sets  of  tail 
beams,  and  with  a  given  rate  of  deflection.  (See  Arts.  393, 
396  and  402.) 


.  —Example.  —  Under  this  rule,  what  should  be  the 
breadth  of  a  Georgia  pine  carriage  beam  of  average  quality, 
20  feet  long  and  12  inches  deep,  to  carry  two  headers  each 


CARRIAGE   BEAMS   WITH   TWO    HEADERS.  289 

12  feet  long ;  the  headers  so  placed  as  to  leave  an  opening 
6  feet  wide  in  the  middle  of  the  width  of  the  floor  ?  The 
floor  beams  are  set  16  inches  from  centres,  and  are  to  carry 
200  pounds  per  foot  superficial,  with  a  deflection  of  0-04 
of  an  inch  per  foot  lineal. 

l-d'       20-6 
Here  /=2O,    m  =  — —  =  — - —  =  7;    ^=12,     ^=12, 

c  =  i^-,  F==  5900,  /— 200  and  ^  =  0-04;  and  by  formula 
(183.) 

,  200  x  20         r/  .N  ,.... 

*  ;=  5900  x  12°  x  0.04  «*  x  H  x  20')  +  (12  x  7')]  =  7-402 

or  the  breadth  should  be   7§   inches. 


417.  —  Carriage  Beams  with  Two  Headers,  Equidistant 
from  Centre,  and  Two  Sets  of  Tail  Beams,  for  Dwellings 
and  for  First-class  Stores  —  Precise  Rules.  —  If,  in  formula 
(183.),  f=  90  and  r  =  0-03,  we  shall  have 


which  is  a  precise  rule  for  carriage  beams  carrying  two 
headers,  equidistant  from  the  centre,  with  two  sets  of  tail 
beams,  in  a  dwelling  or  assembly  room.  (For  an  example, 
see  Art.  418.)  But  if,  instead,  /=  275  and  ?•  =  0-04, 
then  we  shall  have 


which  is  a  precise  rule  for  carriage  beams  carrying  two 
headers,  equidistant  from  the  centre,  with  two  sets  of  tail 
beams,  in  a  first-class  store.  (See  Arts.,  393,  395  and  402.) 


2QO  RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII. 

4-!  8.  —  Examples.  —  Formulas  (184.)  and  (185.)  are  alike, 
except  in  the  numerical  coefficient.  One  example  will 
therefore  suffice  for  an  exemplification  of  the  two.  Let  it  be 
required  to  show  what,  in  a  dwelling,  should  be  the  breadth 
of  a  carriage  beam,  20  feet  long  and  12  inches  deep,  of 
average  quality  of  spruce,  carrying  two  headers  10  feet 
long  ;  these  headers  being  so  placed  as  to  leave  at  the  middle 
of  the  width  of  the  floor  an  opening  8  feet  wide.  The 
beams  are  to  be  placed  16  inches  from  centres. 

Here  we  have  /=  20,  ;;/  =  6,  g  =.  10,  ^/=  12,  c=  i$ 
and  F=  3500  ;  and  by  formula 


i)]  =  5-225 


or  the  breadth  should  be,  say    $J   inches. 

For  a  first-class  store  this  carnage  beam,  if  of  Georgia 
pine,  would  be  required  to  be  7-  103,  say  7J  inches 
broad.  This  result  is  found  by  eliminating  the  two  con- 
stants 3000  and  3500  in  the  above  and  replacing  them  by 
those  required  by  the  new  conditions,  namely,  6875  and 
5900.  Doing  this,  we  find 

6875   3500 

£=  5-225  x-  -x-  -=7-103 
3   J   3000   5900   ' 


419. — Carriage   fleam   with   Two   Header§,  Equidistant 
from  Centre,  and  One    Set   of  Tail  Beam§— Precise  Rule.— 

In  some  cases  the  wells  or  openings  are  at  the  wall  on  each 
side,  and  the  tail  beams  at  the  middle  of  the  floor.     In  this 
arrangement,   if   /    equals   the   length   of   the   tail   beams, 
\fgj    will  equal  the  load  at  the  end  of  one  header. 
By  Art.  415,     b'  —  Wm,    from  which 

V  =  Wm  = 


CARRIAGE  BEAM   WITH  TWO   HEADERS.  29! 

and  formula  (182.)  becomes 


and  since  (Art.  415)     h  =  t  =  I/,     therefore 

•fckt  =  W 

By  substituting  this  in  the  above, 


which  is  a  precise  rule  for  carriage  beams,  carrying  two 
headers,  equidistant  from  the  centre,  with  one  set  of  tail 
beams,  the  rate  of  deflection  being  given.  (See  Arts.  397, 
398,  402,  409  and  411.) 


420. — Example.— What  should  be  the  breadth  of  a  car- 
riage beam  of  hemlock  of  average  quality,  16  feet  long  and 
ii  inches  deep,  carrying  two  headers,  each  10  feet  long, 
placed  equidistant  from  the  centre  of  the  width  of  the  floor, 
and  having  between  them  one  set  of  tail  beams  6  feet  long? 
The  floor  beams,  placed  1 5  inches  from  centres,  are  to  carry 
100  pounds  per  foot  superficial,  with  a  deflection  of  0-035 
of  an  inch  per  foot  lineal. 

Here  we  have  /=  16,  m  =  5,  g-=  10,  j  =  6,  df=  11, 
c—  i£,  F—  2800,  /=  100  and  r=  0-035  '»  and  by  formula 
(186.) 

b  =  28oox°?i-x1o.035  [(A  X  '* X  I6')+  ('° x 6 x  5)]  =  4-907 
or  the  breadth  should  be   4|   inches. 


RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII. 

421.  —  Carriage  Beams  \viili  Two  Headers,  Equidistant 
from  Centre,  and  One  Set  of  Tail  Beams,  for  Dwellings  and 
for  First-class  Stores—  Precise  Rules.  —  If,  in  formula  (186.), 
f=  90  and  r  =•  0-03,  then  we  shall  have 


which  is  a  precise  rule  for  carriage  beams,  carrying  two 
headers,  with  one  set  of  tail  beams  between  them,  at  the 
middle  of  the  floor,  in  a  dwelling  or  assembly  room. 

For  an  example,  see  Art,  4-22. 

But  if,  instead  of  these,  /=  275  and  r  —  0.04,  we 
shall  have 


which  is  a  precise  rule  for  carriage  beams,  carrying  two 
headers,  with  one  set  of  tail  beams  between  them,  at  the 
middle  of  the  floor,  in  a  first-class  store. 

422.— Example.— Formulas  (187.)  and  (188.)  are  alike, 
except  in  the  numerical  coefficient.  One  example  will 
suffice  to  show  the  application  of  both. 

Take  one  coming  under  formula  (187.),  and  in  which 
/  =  20,  m  =  6,  g  —  10,  j  =8,  d—  12,  £=!•§•  and 
JF=.$$oo.  Then,  by  the  formula, 


=  6-415 


or  the  breadth  should  be   6     inches  full. 


423. — Beam  with  Uniformly  Distributed  and  Three  Con- 
centrated   Loads,   the  Greatest  Strain    being  Outside. — In 

Art.  256,  formula  (96.)  is  a  general  rule  for  this  case,  but 


BEAM  CARRYING  THREE  CONCENTRATED   LOADS.         293 

based  upon  the  resistance  to  rupture.  This  rule  may  be 
modified  so  that  it  shall  be  based  upon  the  resistance  to 
flexure.  To  this  end  let  a,  in  formula  (96.),  be  substituted 

/?/ 
by  its  value  in  formula  (154-),        r~>     and  we  have,  taking 


(189.) 


which  is  a  rule,  based  upon  the  resistance  to  flexure,  for  a 
beam  uniformly  loaded,  and  also  carrying  three  concentrated 
loads,  the  largest  of  which  is  not  between  the  other  two. 


424. — Example. — What  ought  to  be  the  breadth  of  a 
beam  of  Georgia  pine  of  average  quality,  20  feet  long  and 
12  inches  deep,  carrying  an  equally  distributed  load  of  4000 
pounds,  together  with  three  concentrated  loads,  namely,  7000 
pounds  at  7  feet  from  the  right-hand  end,  4000  pounds  at 
7  feet  from  the  left-hand  end,  and  3000  pounds  at  3  feet 
from  the  same  end.  (See  Art.  264.)  Allotting  the  symbols  to 
accord  with  the  arrangement  required  under  rule  (189.),  (the 
largest  strain,  as  in  Fig.  55,  not  between  the  other  two),  we 
have  U—  4000,  A'  =  7000,  B'  =  4000,  O  =  3000,  /=  20, 
m  =  j,  n=i$,  s=7,  z>  =  3,  d=  12  and  /**=  5900,  and  let 
r  =  0-04 ;  and  by  formula  (189.) 

b  =  5900  x4i2'7x  0-04 [(§  x  4000  x  13)  +  (7000  x  13)  +  (4000  x  7)  +. 

(3000X3)]  =  11-020 

or  the  breadth  should  be    1 1    inches. 


294  RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII. 

425.—  Carriage  Beam  with  Three  Headers,  the  Greatest 
Strain  being  at  Outside  Header.—  If,   in  formula  (97.),  (Art. 

JR7 

258),  we  substitute  for   a   its  value,         ~     (form.    154.),  we 


shall  have,  taking  £Z7,  Art.  340 


,    _      mf    r5       /          /  g_    >yi  (2QQ  \ 

ZTV/^/**   L4      '        6   \  /_J  \  / 


which  is  a  rule,  based  upon  the  resistance  to  flexure,  for  car- 
riage beams  carrying  three  headers,  with  two  sets  of  tail 
beams,  so  located  (as  in  Figs.  54  and  55)  that  the  header  at 
which  there  is  the  greatest  strain  shall  not  be  between  the 
other  two  headers. 


426. — Example. — What  should  be  the  breadth  of  a  car- 
riage beam  of  Georgia  pine  of  average  quality,  20  feet  long 
and  12  inches  deep,  carrying  three  headers  15  feet  long, 
two  of  them,  for  a  light-well  6  feet  wide,  located  centrally 
as  to  the  width  of  the  floor,  and  the  third  header,  at  the  side 
of  an  opening  for  a  stairway  3  feet  wide  at  one  of  the 
walls?  The  floor  beams,  placed  15  inches  from  centres,  are 
to  carry  200  pounds  per  superficial  foot,  with  a  deflection 
of  0-04  of  an  inch  per  foot  lineal.  (See  Art.  264.) 

Allotting  the  symbols  as  in  Fig.  55,  we  have  /  =  20, 
»*=7,  «=i3,  *  =  7,  v  =  3i  £=  J5>  </=i2,  r=ii, 
F  —  5900,  f=  200  and  r  =  0-04  ;  and  by  formula  (190.)  we 
have 


X  2OO 


or  the  breadth  should  be   8i    inches. 


CARRIAGE   BEAM   WITH   THREE  HEADERS.  2Q5 

427.—  Carriage  Beam  with  Three  Headers,  the  Greatest 
Strain  being  at  Outside  Header,  for  Dwellings.  —  If,  in  for- 
mula (190),  f—  go  and  r=  0-03,  we  shall  have 


which  is  a  rule,  based  upon  the  resistance  to  flexure,  for  car- 
riage beams  in  dwellings  and  assembly  rooms,  to  carry 
three  headers,  with  two  sets  of  tail  beams,  so  located  that 
(as  in  Fig.  55)  the  header  at  which  there  is  the  greatest  strain 
shall  not  be  between  the  other  two, 
For  an  example,  see  Art.  429. 


428.—  Carriage  Beam  with  Three  Headers,  the  Greatest 
Strain  being   at  Outside  Header,  for  First-elass   Stores.  —  If, 

in  formula  (190.),  f  —  275    and    r  =  0-04,    we  shall  have 


(192.) 


which  is  a  rule,  based  upon  the  resistance  to  flexure,  for  car- 
riage beams  in  first-class  stores,  to  carry  three  headers,  with 
two  sets  of  tail  beams,  so  located  that  (as  in  Fig.  55)  the 
header  at  which  there  is  the  greatest  strain  shall  not  be 
located  between  the  other  two. 


429.— Examples. — Formulas  (191.)  and  (192)  are  alike, 
except  in  the  numerical  coefficient,  which,  in  the  rule  for 
dwellings  and  assembly  rooms,  is  3000,  while  for  first-class 
stores  it  is  6875.  An  example  under  one  rule  will  serve  to 
illustrate  the  other,  by  a  simple  substitution  of  the  proper 
coefficient. 

As  an    example   under  rule  (191)  :  What  should  be  the 


296  RESISTANCE  TO  FLEXURE  —  FLOOR  BEAMS.     CHAP.  XVII. 

breadth,  in  a  dwelling,  of  a  carriage  beam  of  white  pine  of 
average  quality,  20  feet  long  and  12  inches  deep,  carrying 
three  headers  12  feet  long,  so  placed  as  to  provide  an  open- 
ing 4  feet  wide  for  a  stairs  at  one  wall,  and  a  light-well  6 
feet  wide  at  the  middle  of  the  width  of  the  floor?  The 
floor  beams  are  placed  16  inches  from  centres.  (See  Art. 
264.) 

Allotting  the  symbols  as  in  Fig.  55  we  have  /  =  20,  m  —  7, 
n  =  13,  s=  7,  v  =  4,  £-—12,  d  =  12,  c  =  ij  and  .F=  2900; 
and  by  formula  (191.) 


b  =  -^v*,  v  Ki  x  i^  x  13  x  20)+  12  (7^13  +  73-4')]  =  8-052 

2QOO  X  1  ^ 

or  the  breadth  should  be   8    inches. 

This  is  the  breadth  when  of  white  pine,  and  in  a  dwelling. 
If,  instead,  it  be  required  of  Georgia  pine,  and  for  a  first- 
class  store,  then  the  breadth  just  obtained,  treated  by  the 
proper  constant  and  numerical  coefficient,  and  at  the  same 
time  relieved  from  those  applying  to  the  previous  case, 
will  be 

6875        200O 

b  —  8-CX2  x  —  —  x  -    -  =  0-070 
3000     5900 

or  the  breadth,  when   of  Georgia  pine,  and  for  a  first-class 
store,  should  be   9^   inches. 


4-30.—  Beam*  with  Uniformly  Distributed  and  Three 
Concentrated  Loads,  the  Greatest  Strain  being  at  middle 
Load. — In  Art.  262  a  rule  is  given  for  beams  uniformly 
loaded,  and  also  carrying  three  concentrated  loads,  the  mid- 
dle one  of  which  produces  the  greatest  strain.  This  rule  is 
based  upon  the  resistance  to  rupture.  It  may  be  modified  to 


BEAM   WITH   THREE   CONCENTRATED   LOADS.  2Q/ 

depend  upon  the  resistance  to  flexure  by  substituting,  in  for- 

/?/ 
mula  (99.),  for  a  its  value   -=j-   (form.  1&4-),  (taking  | U,  Art. 

340),  thus 

(193.) 


which  is  a  rule,  based  upon  resistance  to  flexure,  for  beams 
carrying  a  uniform  load  (U)  and  three  concentrated  loads 
(A,  B'  and  C),  the  middle  one  of  which  produces  the 
greatest  strain  of  the  three,  as  in  Fig.  56. 

431. — Example. — What  should  be  the  breadth  of  a  beam 
of  Georgia  pine  of  average  quality,  20  feet  long  and  14 
inches  deep,  and  carrying  4000  pounds  uniformly  distrib- 
uted, 6000  pounds  at  4  feet  from  one  end,  6000  pounds  at 
9  feet  from  the  same  end,  and  7000  pounds  at  6  feet  from 
the  other  end  ;  with  a  deflection  of  0-04  of  an  inch  per  lineal 
foot?  (See  Art.  264.) 

Assigning  the  symbols  as  per  figure,  we  have,  £7  =  4000, 
A  =  6000,  B'  =  7000,  C  =  6000,  1=20,  m  =  9,  »=n, 
s  =  6,  v  =  4,  */=  14,  F  =  5900  and  r  =  0-04  ;  and  by  for- 
mula (19  3. \ 


(6000  x  1 1  x  4)]  =  9- 163 
or  the  breadth  should  be   9^   inches. 

432. — Carriage  Beam  with  Three  Headers  the  Oreate§t 
Strain  being  at  Middle  Header.— If,  in  formula  (106.),  (Art. 

/?/ 
267),  there  be  substituted  for   a    its  value        -   (form. 

we  shall  have,  taking  f  £7,  Art.  340, 
b  =      ~  m 


298  RESISTANCE  TO  FLEXURE — FLOOR  BEAMS.     CHAP.  XVII. 

which  is  a  rule,  based  upon  resistance  to  flexure,  for  carriage 
beams  carrying  three  headers  and  two  sets  of  tail  beams, 
so  placed  (as  in  Fig.  56)  that  of  the  strains  produced  at  the 
headers,  the  greatest  shall  be  at  the  header  which  is  between 
the  other  two. 


4-33.—  Example.  —  What  should  be  the  breadth  of  a  car- 
riage beam,  20  feet  long  and  12  inches  deep,  of  Georgia 
pine  of  average  quality,  carrying  three  headers  14  feet  long, 
so  placed  as  to  provide  a  stair  opening  4  feet  wide  at  one 
wall,  and  a  light-well  5  feet  wide  6  feet  from  the  other 
wall  ?  The  floor  beams,  placed  15  inches  from  centres,  are 
to  carry  200  pounds  per  foot  superficial,  with  a  deflection 
of  0-04  of  an  inch  per  foot  lineal.  (See  Art.  264.) 

Assigning  the  symbols  as  per  Fig.  56  we  have,  /  =  20, 
m  =  9,  n  =  1  1,  s  =  6,  v  =  4,  g  =  14,  d  =  12,  <?,==  if, 
F  =  5900,  f=  200  and  r  =  0-04;  and  by  formula 


2OO 


•  _         _ 

=  S900xi2'xo.o4L9(*XI*  x  11  x  20  +  14  x  6')  -.- 

(14  x  ii  xy'—  42)]  =  8-651 
or  the  breadth  should  be,  say   8g    inches. 

434.  —  C?arr5age  CScaiai  with  Three  Headers,  the  Greatest 
Strain  being  at  Middle  Header,  for  Dwellings.  —  If,  in  for- 
mula (194-),  f=  90  and  r  —  0-03,  we  shall  have 


b  =  {,m(lcnl+gs^+gn  (m*-v*)]      (195.) 

which  is  a  rule,  based  on  resistance  to  flexure,  for  carriage 
beams  in  dwellings  and  assembly  rooms,  to  carry  three 
headers,  with  two  sets  of  tail  beams  relatively  placed  as  in 
Fig.  56,  so  that,  of  the  three  strains  produced  at  the  headers, 
the  greatest  shall  be  at  the  header  which  is  between  the 
other  two.  (For  an  example,  see  Art.  436.) 


CARRIAGE   BEAM   WITH    THREE    HEADERS.  299 

4-35.  —  Carriage  Beam  with  Three  Headers,  the  Greatest 
Strain  being  at  middle   Header,    for  First-class  Stores.  —  If, 

in  formula   (194*),   f=  275     and    r  —  0-04,    then  we  shall 
have 


b  =  \m  (Icnl+gs*}  +gn  K-tf)]      (196.) 

which  is  a  rule,  based  on  resistance  to  flexure,  for  carriage 
beams  in  first-class  stores,  to  carry  three  headers,  with  two 
sets  of  tail  beams  relatively  placed  as  in  Fig.  56,  so  that,  of 
the  three  strains  produced  at  the  headers,  the  greatest  shall 
be  at  the  header  which  is  between  the  other  two. 


4-36.  —  Example.  —  Formulas  (193.)  and  (106.)  being  alike, 
except  in  the  numerical  coefficient,  a  single  example  will 
suffice  to  illustrate  them. 

In  a  dwelling,  what  should  be  the  breadth  of  a  carriage 
beam  of  oak  of  average  quality,  20  feet  long  and  12  inches 
deep,  to  carry  three  headers  15  feet  long,  with  two  sets  of 
tail  beams,  so  placed  as  to  provide  a  stair  opening  4  feet 
wide  at  one  wall,  and  a  light-well  7  feet  wide,  distant  5 
feet  from  the  other  wall  ?  The  beams  are  to  be  placed  1  5 
inches  from  centres.  (See  Art.  264.) 

Arranging  the  symbols  in  the  order  in  which  they  appear 
in  Fig.  56,  we  have,  /  =  20,  m  =  8,  n  =  12,  s  =  5,  v  =  4, 
£•=15,  d  =  12,  c=  i£  and  F—  3100;  and,  by  formula 


or  the  breadth  should  be,  say   8J   inches. 


300  RESISTANCE  TO  FLEXURE— FLOOR  BEAMS.     CHAP.  XVII. 


QUESTIONS   FOR   PRACTICE. 


437. — In  a  dwelling:  What  should  be  the  depth  of  white 
pine  beams  of  average  quality;  they  being  18  feet  long  and 
3  inches  broad,  placed  18  inches  from  centres,  and  allow- 
ed to  deflect  0-03  of  an  inch  per  foot? 

438. — In  a  first-class  store :  What  should  be  the  breadth 
of  the  floor  beams  of  spruce  of  average  quality,  19  feet 
long,  13  inches  deep,  placed  13  inches  from  centres,  and 
with  a  deflection  of  0-04  of  an  inch  per  foot  ? 

439. — In  a  dwelling:  What  ought  to  be  the  breadth  of  a 
header  of  white  pine  of  average  quality,  14  feet  long  and 
13  inches  deep,  carrying  one  end  of  a  set  of  tail  beams  15 
feet  long,  and  with  a  rate  of  deflection  of  0-03  of  an  inch 
per  foot  ? 

440. — In  the  floor  of  an  assembly  room,  in  which  the 
beams  are  15  inches  from  centres:  What  should  be  the 
breadth  of  a  carriage  beam  of  spruce  of  average  quality, 
20  feet  long  and  12  inches  deep,  carrying  one  header  13 
feet  long,  located  at  5  feet  from  open  end  ?  The  deflection 
allowable  is  0-03  of  an  inch  per  foot. 

441.— In  the  floor  of  a  first-class  store,  where  the  beams 
are  15  inches  deep  and  set  14  inches  from  centres :  What 
should  be  the  breadth  of  a  carnage  beam  24  feet  long, 
of  Georgia  pine  of  average  quality,  carrying  two  headers 


QUESTIONS   FOR  PRACTICE.  3OI 

1 6  feet  long,  located,  one  at  9  feet  from  one  end,  and  the 
other  at  7  feet  from  the  other  end,  with  two  sets  of  tail 
beams?  The  deflection  is  0-04  of  an  inch  per  foot. 

4-4-2. — In  the  floor  of  a  first-class  store,  with  beams  16 
inches  deep  placed  15  inches  from  centres  :  What  should  be 
the  breadth  of  a  carriage  beam  of  Georgia  pine  of  average 
quality,  26  feet  long,  and  carrying  three  headers  18  feet 
long,  located  as  in  Fig.  54,  one  at  4  feet  from  one  wall, 
another  at  8  feet  from  the  same  wall,  and  the  third  at  8 
feet  from  the  other  wall  ?  The  deflection  to  be  0-04  of  an 
inch  per  foot. 


CHAPTER    XVIII. 


BRIDGING     FLOOR    BEAMS.* 


ART.  443.— Bridging  Defined. — Bridging  is  a  system  of 
bracing  floor  beams.  Small  struts  are  cut  to  fit  between 
each  pair  of  beams,  and  secured  by  nails  or  spikes ;  as 
shown  in  Fig.  65.  The  effect  of  this  bracing  is  decidedly 


FIG.  65. 

beneficial  in  sustaining  any  concentrated  weight  upon  a  floor. 
The  beam  immediately  beneath  the  weight  is  materially  as- 

*  The  principles  upon  which  this  chapter  is  based  the  author  first  made 
public  in  an  article  which  appeared  in  the  Scientific  American,  July  a6th,  1873, 
entitled  "On  Girders  and  Floor  Beams — The  Effect  of  Bridging." 


EXPERIMENTAL   TEST.  303 

sisted,  through  these  braces,  by  the  beams  on  each  side  of  it. 
It  is  customary  to  insert  rows  of  cross-bridging  at  every  five 
to  eight  feet  in  the  length  of  the  beams. 

It  is  the  usual  practice,  where  the  ceiling  of  a  room  is 
plastered,  to  attach  the  plastering  laths  to  cross-furring,  or 
narrow  strips  of  boards  crossing  the  beams  at  right  angles, 
and  nailed  to  their  bottom  edge.  These  strips  are  set  at, 
say  12  inches  from  centres,  and  when  firmly  nailed  to  the 
beams  act  as  a  tie  to  sustain  the  lateral  thrust  of  the  bridg- 
ing struts.  The  floor  plank  at  the  top  serve  a  like  purpose. 


.— Experimental  Test. — To  test  the  effect  of  bridging, 
about  three  years  since  I  constructed  a  model,  and  sub- 
jected it  to  pressure.  It  was  made  upon  a  scale  of  1%  inches 
to  the  foot,  or  \  of  full  size,  and  represented  a  floor  of  seven 
beams  placed  16  inches  from  centres,  each  beam  being 
3  x  10  inches  and  14!  feet  long.  These  beams  were  con- 
nected by  two  rows  of  cross-bridging,  and  secured  against 
lateral  movement  by  strips  representing  floor  plank  and  ceil- 
ing boards,  which  were  nailed  on  top  and  beneath.  There 
were  four  strips  at  each  row  of  bridging,  two  above  and  two 
below. 

Before  putting  these  beams  in  position  in  the  model,  I 
submitted  each  beam  to  a  separate  test,  and  ascertained  that 
to  deflect  it  one  tenth  of  an  inch  required  from  37  to  40 
pounds,  or  on  the  average  38^  pounds. 

With  the  model  completed,  the  beams  being  bridged,  it 
required  a  pressure  of  1554  pounds  applied  at  the  centre  of 
the  middle  beam  to  deflect  it  as  before,  one  tenth  of  an  inch. 
And  while  this  pressure  deflected  the  central  beam  to  this 
extent,  the  beam  next  adjoining  on  each  side  was  deflected 
0-0808  of  an  inch,  the  ones  next  adjoining  these  were  each 
deflected  0-0617  of  an  inch,  while  the  two  outside  beams 


3°4  BRIDGING   FLOOR   BEAMS.  CHAP.    XVIII. 

were  each  depressed  0-0425  of  an  inch.  Had  there  been 
more  than  seven  beams,  and  all  bridged  together,  the  effect 
would  doubtless  have  been  still  better. 

As  the  result  of  this  test  of  the  effect  of  bridging,  we  have 
one  beam  sustaining  155!  pounds  with  the  same  deflection 
that  was  produced  by  38^  pounds  before  bridging,  or  an  in- 
crease of  H7TV  pounds;  an  addition  of  more  than  three 
times  the  amount  borne  by  the  unbridged  beam. 


445. — Bridging— Principle§  of  Resistance. — The  assist- 
ance contributed  by  the  adjacent  beams  to  a  beam  under 
pressure  may  be  computed,  but  preliminary  thereto  we  have 
these  considerations,  namely  : 

First. — The  deflections  of  a  beam  are  (within  the  limits 
of  elasticity)  in  proportion  to  the  weights  producing  the  de- 
flections. Thus,  if  one  hundred  pounds  deflect  a  beam  one 
tenth  of  an  inch,  two  hundred  pounds  will  deflect  it  two 
tenths  of  an  inch.  From  which,  knowing  the  deflection  of  a 
beam,  we  can  compute  the  resistance  it  offers. 

Second. — The  resistance  thus  offered,  being  at  a  distance 
from  the  beam  suffering  the  direct  pressure,  is  not  so  effect- 
ual as  it  would  be  were  it  in  direct  opposition  to  the  pres- 
sure. It  is  diminished  in  proportion  to  its  distance  from 
that  beam. 


446. — Resistance  of  a  Bridged  Beam. — Based  upon  the 
two  preceding  considerations,  we  will  construct  a  rule  by 
which  to  measure  the  increase  of  resistance  derived  from 
the  adjacent  beams  through  their  connection  by  cross- 
bridging. 

Let  Fig.  66  represent  the  cross-section   of  a  tier  of  floor 


INCREASED    RESISTANCE   OF   BRIDGING. 


305 


beams  connected  by  cross-bridging,  in  which  C  is  the  lo- 
cation of  a  concentrated  weight,  AB  the  distance  on  one 
side  of  the  weight  to  which  the  deflecting  influence  acting 


FIG.  66. 


through  the  cross-bridging  is  extended,  BC  the  deflection 
at  the  weight,  and  DE  the  deflection  of  one  of  the  beams 
E,  caused  by  the  weight  at  C.  The  triangles  ABC  and 
ADE  are  similar,  and  tkeir  sides  are  in  proportion.  Put- 


ting 
DE, 


p     for     AB, 
we  have 

AB 

P 


m    for    AD,     a'    for    BC,    and     V    for 


BC  :  :    AD  :  DE 


=  •? 


This  is  the  measure  of  the  deflection  at  E,  or  at  any  one  of 
the  beams  the  distance  of  which  from  A  is  equal  to  m, 
and,  since  the  deflections  are  as  the  weights  producing 
them,  therefore  b ',  the  deflection  at  E,  measures  the 
strain  there,  when  a'  measures  that  at  C. 

It  is  required,  however,  to  know  not  only  the  resistance 
offered  by  each  beam,  but  also  what  weight  r,  acting  at 
C,  would  be  required  to  overcome  this  resistance.  The 
line  AB  (or  /)  may  be  considered  to  serve  as  a  lever,  hav- 
ing its  fulcrum  at  A.  The  weight  r,  at  B,  acting  in 
the  line  BC,  is  opposed  at  D  by  b' ' ,  the  resistance  of 
the  beam  at  E,  acting  in  the  line  ED,  with  the  leverage 
m.  The  weight  r  will  act  with  the  moment  rp,  and  I' 
will  resist  with  the  moment  b'm.  Putting  these  moments 

in  equilibrium,  we  have    b'm  —  rp,    or    r  =  b'—. 


306  BRIDGING  FLOOR  BEAMS.  CHAP.   XVIII. 

In  this,  substituting  for    b'    its  value  as  above  found,  we 
have 

,m    m 

r  =  a  —  x—  • 
P    P 


This  weight  r  represents  the  effect  at  C  of  the  resist- 
ance to  deflection  of  any  beam  whose  distance  from  A  is 
equal  to  m,  and  where  a'  equals  the  load  borne  by  the 
beam  at  B,  and  /  is  put  for  the  distance  AB. 


44-7.  —  Summing  tlie  Resistances.  —  Let  the  distance  from 
centres  between  the  floor  beams  be  represented  by  c,  and 
the  number  of  spaces  from  A  to  any  beam,  as,  for  example, 
that  at  D,  by  n  ;  then  m  =  nct  and  substituting  this  value 
for  m  in  (197.)  we  have 


r  =  n*~  (198.) 

In  this  expression,  a',  c*  and  p3  are  constants,  or  quanti- 
ties which  remain  constant  for  the  several  values  of  r 
which  are  to  be  obtained  from  the  resistances  of  the  several 

beam?.     For  convenience,  put    /    for    —  j     and  then 

r  =  vtt  (199.) 

With  this  expression,  the  various  values  of  r  may  be  ob- 
tained and  grouped  together.  In  doing  this,  we  have,  for 
the  first  beam  from  A,  n  =  i  ;  for  the  second,  n  =  2;  for 


SUM   OF   INCREASED    RESISTANCES.  307 

the  third,  n  —  3,  and  so  on  to  the  middle  or  point  of  great- 
est depression.  Therefore  the  whole  resistance  will  be 

R'  =  t  +  2V  +  3V  +  4V  +  etc. 
R'  —  t  (i  +  4  +  9  +  16  +  etc.) 

This  gives  the  resistance  on  one  side  of  the  point  C.  The 
beams  on  the  other  side  afford  a  like  resistance  ;  and  the 
sum  of  the  two  resistances  will  be 

R  =  2t  (i  +  4  +  9  +  16  +  etc.) 
R  =  2  —T-  (i  +  4  +  9  +  16  +  etc.) 


448.  —  Example.  —  When  a  concentrated  weight  deflects 
six  beams  on  each  side  of  it,  they  being  placed  16  inches 
from  centres  :  What  will  be  the  amount  of  resistance  to  de- 
flection offered  by  the  twelve  beams,  the  beam  upon  which 
the  weight  rests  being  capable  of  sustaining  alone,  unaided 
by  the  adjoining  beams,  1000  pounds  ? 

Here  a'  =  1000,  c  =  i%  and  p  =  ?xi$  =  9^.  There- 
fore, by  formula  (200.}, 


2  X  IOOO  X 

=  37H-3 


This  3714  pounds  is  the  resistance  offered  by  the  twelve 
beams,  through  the  means  of  bridging,  and  is  nearly  four 
times  the  amount  that  the  centre  beam,  unaided  by  the 
bridging,  would  carry  with  a  like  deflection.  The  combined 
resistance  of  all  the  beams  would  be  3714+1000  =  4714 
pounds. 


308  BRIDGING   FLOOR   BEAMS.  CHAP.    XVIII. 

44-9-  —  Assistance  Derived  from  Cross-bridging.  —  Just 
how  many  beams  on  each  side  will  be  affected,  and  by  their 
resistance  contribute  in  aiding  the  beam  at  £7,  will  depend 
upon  circumstances.  The  bridging  will  be  effective  in  re- 
sisting deflection  in  proportion  to  the  elevation  of  the  angle 
at  which  the  bridging  pieces  are  placed,  which  will  be 
directly  as  the  depth  of  the  beams  and  inversely  as  their 
distance  apart.  It  will  also  be  in  proportion  to  the  faithful- 
ness with  which  the  work  of  bridging  is  executed.  From 
these  considerations,  and  from  the  experiment  of  Art.  44-4, 
we  conclude  that,  in  well-executed  work,  we  shall  have 

d 


An  equally  distributed  load  upon  a  floor  beam  is  represented 
(Art.  92)  by  cfl.  A  load  at  the  centre  of  the  beam  produc- 
ing an  equal  effect  will  be  f  of  this,  or  \cfl.  The  symbol 
a'  (form.  200.)  represents  the  load  at  the  middle  of  a  floor 
beam,  and  therefore 

a'  =  fc/7 

These  values  of  /  and  a'  may  be  substituted  for  these 
symbols  in  formula  (®00.\  and  the  result  will  be 


R=  2  -TT-j  (i  +  4  +  9  +  etc.)  or, 


(901.. 


In  this  rule  R  equals  the  additional  resistance  to  a  concen- 
trated weight  on  a  beam,  obtained  from  adjacent  beams 
through  the  cross-bridging. 


USEFUL   FOR   CONCENTRATED    WEIGHTS.  309 

450. —  Xumfoer    of   Beams    Affording    Assistance. —  The 

value  of  /,    as  above,    is   -.     The  symbol  -n   being  put  for 

the  number  of  spaces  on  each  side  of  the  beam  sustaining  the 
concentrated  weight,  over  which  this  weight  exerts  an  in- 
fluence;  or  /,  the  distance  AB  of  Fig.  66;  and  c  for  the 
distance  apart  from  centres  at  which  the  beams  are  placed  ; 

then,    /  =  —  =  nc  ;  from  which  we  have 

n  =  %  (®02.) 


To  apply  this  rule  :    How   many    beams  on  each  side  of  a 

concentrated  weight  would  contribute  towards  sustaining  it, 

when  they  are    12    inches  deep,  and   16  inches  from  centres? 

Here  we  have     d  =  12     and     c  =  i-J-,     and  therefore 

n  =  -p  =  6f    say  7  spaces. 

3 

In  seven  spaces,  six  beams  will  be  affected. 


451.  —  Bridging  Useful  in  Sustaining  Concentrated 
Weights. — The  results  shown  in  Art.  448  illustrate  the  ad- 
vantage of  cross-bridging  in  resisting  concentrated  weights, 
and  show  the  importance  of  always  having  floor  beams 
bridged,  and  the  work  faithfully  executed.  The  advantage, 
however,  of  cross-bridging  inheres  only  in  the  case  of  concen- 
trated weights.  For,  although  in  the  example  of  Art.  44-8,  the 
13  beams  sustained  by  their  united  resistance  a  concentrated 
weight  of  4714  pounds,  yet  it  will  be  observed  that  this  is 
not  the  limit  of  their  power,  for  they  are  each  capable  of 
sustaining  1000  pounds  placed  at  the  middle,  or  together, 
13,000  pounds;  nearly  three  times  the  previous  amount. 


310  BRIDGING   FLOOR   BEAMS.  CHAP.    XVIII. 

4-52. — I5icrea§ed  Resistance  Due  to  Bridging. — A  useful 
application  of  the  results  of  this  investigation  is  found  in 
determining  the  amount  of  concentrated  weight  which  may 
be  borne  upon  a  floor  beam.  As  an  example  :  In  a  dwelling 
with  well-bridged  floor  beams  of  an  average  quality  of 
white  pine,  3  x  10  inches,  and  16  feet  long,  what  concen- 
trated weight  may  be  safely  sustained  at  the  middle  of  one 
of  them  ? 

The  distance  from  centres  at  which  these  beams  should 
be  placed  is  had  by  formula  (144-),  Art.  362, 


ibd3       i  -55 


the  value  of    i    being  taken  as  found  in  Art.  361. 

With  the  above  value,  c  —  1-135,  we  may,  by  formula 
(202.),  find  the  distance  to  which  the  effect  of  the  weight 
extends  on  each  side,  thus : 


io  io 


say  8  spaces,  or  7  beams.  The  symbols  of  formula  (20 l.\ 
applied  in  this  case,  will  be  as  follows  :  c  =  i  •  135,  /  =  90, 
/=  16  and  d  =  io,  and  the  squares  in  the  parenthesis 
extend  to  7  places.  Therefore 


n      5xi-  i35xgox  16  , 

R  4  x  IQ^"     -(1+4+9+16  +  254-36 


=  18  x  i-"i35~ftx  140 


EXAMPLE,   BY    LOGARITHMS.  311 

The  product  of  these  factors,  one  of  them  being  raised  to 
a  high  power,  will  best  be  obtained  by  logarithms,  thus  : 

Log.   1-135  =  0-0549959 

5 


0-2749795 

Log.  1  8     =  1-2552725 
Log.  140    =  2-1461280 

4746-6   =  3-6763800 

The  product  of  the  factors,  or  the  value  of  R,  is  there- 
fore equal  to  4746-6  pounds.  This  is  the  increased  resist- 
ance. The  resistance  offered  by  the  beam  upon  which  the 
weight  is  laid  equals  (Art.  449) 


To  this,  adding  the  increase  =  4746-6 
we  have        5768-  1 

as  the  total  resistance  to  a  concentrated  load  at  the  middle 
of  the  beam,  when  assisted  by  7  beams  on  each  side  by 
cross-bridging. 


CHAPTER    XIX. 


ROLLED-IRON      BEAMS. 

ART.  453. — Iron  a  Substitute  for  Wood. — When  the 
beams  composing  a  floor  are  of  wood,  they  are  of  rectangular 
form  in  cross-section.  Investigations  into  the  philosophy  of 
the  transverse  strain,  by  which  the  importance  of  depth  was 
developed,  led  to  the  use  of  beams  of  which  the  rectangle 
of  cross-section  was  narrow  and  high.  Owing  to  the  liabil- 
ity, in  wooden  beams  as  generally  used,  of  destruction  by 
conflagration  and  by  other  causes,  iron  was  introduced  as 
a  substitute.  The  greater  cost  of  this  material  over  that 
of  wood,  made  it  important,  now  more  than  ever,  to  give  to 
the  beam  that  shape  which  should  prove  the  strongest. 

454. — iron    Beam— Its    Progressive    Development. — In 

the  use  of  iron  as  a  floor  beam,  economical  considerations 

reduced  the  breadth  until  the 
beam  became  weak  laterally.  To 
remedy  this  defect,  metal  was 
added  at  the  top  and  bottom  in 
the  form  of  horizontal  plates, 
and  these  were  connected  to  the 
thin  vertical  beam  by  angle  irons 
as  in  Fig.  67  ;  the  whole  forming 
what  is  known  as  the  plate  beam 
or  girder.  This  expedient  served 
FlG-  67-  not  only  to  stiffen  the  thin 

vertical  beam  laterally,  but   added   very   greatly  to  its  ab- 


PROPORTIONS   BETWEEN   FLANGES   AND   WEB.  313 

solute  strength.  The  added  material  had  been  placed 
just  where  it  would  do  the  greatest  possible  good  ;  at  a 
point  far  removed  from  the  neutral  axis  of  the  beam. 

455. — Rolled-Iron  Beam— Its  Introduction. — The  in- 
crease of  strength  obtained  in  the  plate  beam  (Fig.  67)  was  so 
great  that  it  became  popular.  To  supply  the  demand,  iron 
manufacturers,  at  great  expense,  made  rolls  similar  to  those 
for  making  railroad  iron,  by  which  they  were  enabled  to  fur- 
nish beams  (Fig.  68)  rolled  out  in  one  piece,  with  all  the  best 
features  of  the  plate  beam,  and  which  could  be  much  more 
readily  and  cheaply  made.  Owing  to  the 
large  cost  of  the  rolls,  only  a  very  few 
sizes  were  at  first  made,  but  these  few 
only  increased  the  demand.  The  man- 
ufacturer, thus  encouraged,  made  rolls 
for  other  sizes,  and  thus  the  number  of 
beams  was  increased,  until  now  we 
have  them  in  great  variety,  from  4  to 
15  inches  high.*  FlG-  68- 

4-56. — Proportions  between  Flanges  and  Web. — These 
beams,  as  usually  made,  have  the  top  and  bottom  plates,  or 
flanges,  of  the  same  form  and  size.  In  wrought-iron  the 
resistances  to  rupture,  by  compression  and  by  tension,  are 
not  equal.  When  the  load  upon  the  beam,  however,  is  not 
so  large  as  to  strain  the  metal  beyond  the  limits  of  elasticity, 

*  There  were  exhibited  at  the  Centennial  Exposition  at  Philadelphia,  by  the 
Union  Iron  Co.,  of  Buffalo,  a  15  inch  beam  52  feet  in  length,  and  a  9  inch 
beam  80  feet  long.  This  is  believed  to  be  the  limit  reached  in  American 
manufacture  at  the  present  time.  The  English  and  Germans,  however,  are  roll- 
ing them  larger.  A  German  exhibit  in  Machinery  Hall  contained  beams  from 
Burbach  half  a  metre  (19-69  inches)  high  by  15  metres  (49-21  feet)  long. 


3H  ROLLED-IRON  BEAMS.         CHAP.  XIX. 

then  it  resists  both  compression  and  tension  equally  well,  and 
hence  the  propriety  of  having-  the  top  and  bottom  flanges 
equally  large. 

The  manifest  advantage  of  having  the  material  accumu- 
lated at  a  distance  from  the  neutral  axis,  has  led  to  putting 
as  much  as  possible  of  the  area  of  the  whole  section  into  the 
flanges,  and  thereby  reducing  the  web  or  vertical  part  to  the 
smallest  practical  thickness.  The  web  is  required  to  main- 
tain the  connection  between  the  top  and  bottom  flanges,  and 
to  resist  the  shearing  effects  of  the  load.  In  rolled-iron 
beams,  as  usually  made,  the  thickness  of  the  web  is  more 
than  sufficient  to  resist  these  strains. 


457. — The  Moment  of  Inertia  Arithmetically  Considered. 

— For  the  intelligent  use  of  the  rolled-iron  beam  as  a  substi- 
tute for  the  wooden  beam  in  floors,  as  well  as  for  other  uses, 
the  rules  already  given  need  modification. 

The  resistance  of  a  beam  to  flexure  or  bending  is  termed 
its  moment  of  inertia.  This  is  represented  in  symbolic  for- 
mulas by  the  letter  7.  In  formula  (111.),  (Art.  300),  the 
coefficient  -£%  and  the  symbols  bd3  represent  the  moment 
of  inertia,  and  /,  its  symbol,  may  be  substituted  for  them, 

thus: 

PNS        PN3 


6  = 


rl 


The  moment  of  inertia  for  any  cross-section  is  equal  to 
the  sum  of  the  products  of  each  particle  of  the  area  o£  the 
cross-section,  into  the  square  of  its  distance  from  the  neutral 
axis.**  For  example :  in  a  beam  with  a  cross-section  of  the 
I  form,  a  horizontal  line  drawn  through  the  centre  of  area 
of  the  cross-section  will  be  the  neutral  line  for  strains  within 

*  Rankine's  Applied  Mechanics,  Art.  573. 


MOMENT   OF   INERTIA. 


315 


the  limits  of  elasticity.  Let  the  area  be  divided  into  a  large 
number  of  small  areas.  Then,  for  the  portion  of  the  figure 
above  the  neutral  line,  multiply  each  of  these  small  areas  by 
the  square  of  its  vertical  distance  above  the  neutral  line,  and 
the  sum  of  these  products  will  equal  the  moment  of  inertia 
for  the  upper  half  of  the  section.  A  like  process  will  give 
the  moment  of  inertia  for  the  lower  half.  The  two  in  this 
case  will  be  equal,  and  their  sum  is  the  moment  of  inertia 
for  the  whole  section.  The  result  thus  obtained  will  not  be 
exact,  but  will  approach  accuracy  in  proportion  to  the  small- 
ness  of  the  parts  into  which  the  area  of  the  cross-section  is 
divided. 


458. — Example  A. — As  an  illustration,  let  A  BCD,  in  Fig. 
69,  represent  the  cross-section  of  a  beam ;  MN,  drawn 
through  the  middle  of  the  height  AD,  being  -  _  . 
the  neutral  axis  ;  and  let  the  lines  EF,  GH,  IJ, 
KL,  OP,  QR,  and  ST  divide  the  area  ABMN 
into  twenty  equal*  parts.  The  four  squares  in 
each  horizontal  row  are  equally  distant  from 
the  neutral  line  MN,  and  may  therefore  M 
be  taken  together.  Suppose  each  of  these 
squares  to  measure  2x2  inches,  then  the  area 
of  each  will  be  4  inches,  and  of  the  four  in 
each  horizontal  row  will  be  4x4=16  inches  D 
area.  The  distances  from  the  neutral  line  to 
the  centre  of  each  square  will  be  as  follows  : 

In  the  first    row,  D  =  i 

"     "  second  "  D  =  3 

"     "  third       "  D  —  5 

"     "  fourth     "  D  =  7 

"     "  fifth         "  D  =  9 


— 





P       R       T 

FIG.  69. 


3*6  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

Their  moment  of  inertia  will  be  as  follows  : 

In  the  first   row,  7,  =  16  x  i2  =  16  x    i 

"     "  second  "  I2  =  16  x  3'  =  16  x    9 

"     "  third      "  I3  —  16  x  52  =  16  x  25 

"     "  fourth    "  74  =  16  x  f  =  16  x  49 

"     "fifth        "  75  =  16  x  92  =  16  x  81 

and  their  sum     7=  16(1  +  9  +  25  +  49  +  81)=  16  x  165  =  2640. 


459. — Example  B. — If  we  subdivide  each  of  the  squares 
in  Fig.  69,  and  take  the  sum  of  the  products  as  before,  the 
result  will  be  larger  and  nearer  the  truth.  For  example : 
divide  each  of  the  squares  into  four  equal  parts,  each  one 
inch  square.  There  will  be  eight  of  these  parts  in  a  row, 
and  ten  rows.  The  area  of  each  row  will  be  8x  1=8,  and 
their  distances  from  the  neutral  line  will  be  -J,  f,  f ,  J,  f , 
V-»  ~/>  V->  ¥  and  V-  respectively.  The  moments  of 
inertia  will  be  as  follows : 

In    the  first    row,  /,  =  8  x    (£)a  =  8  x      l  —  2x      i 

"  second   "  7,  =  8  x    (f)2  :=8x      f  =  2  x      9 

"  third       "  /,  =  8  x    (f)2  =  8  x    *£.  =  2  x    25 

"  fourth     "  74  =  8  x    Q-)2  =  8  x    **-  =  2  x    49 

"  fifth         "  7,  =  8x    (|)*  =  8x    ^L=2x    81 

"  sixth       "  I6  =  8  x  (^  =  8  x  -LfL  =  2  x  121 

"  seventh  "  77  =  8  x  (-V3-)'  =  8  x  .IJA  =  2  x  169 

"  eighth    u  /,  =  8  x  (iff  =  8  x  AjA  =  2  x  225 

"  ninth       "  /.  =  8  x  (-1/)2  =  8  x  i fi  =  2  x  289 

"  tenth     N  "  •  7/0  =  8  x  (J/)2  =  8  x  AJJ.  —  2  x  361 

which  is  equal  to  twice  the  sum  of  the  series  of 

1+9+25+  etc. 

or,  7=2x1330=2660 

This  result  exceeds  in  amount  the  previous  one    (2640). 


MOMENT   OF   INERTIA   COMPUTED.  317 

460. — Example  C. — If  the  eighty  squares  of  this  last 
trial  be  each  subdivided  into  four  equal  parts,  the  whole 
cross-section  will  contain  4x80=320  parts;  there  will 
be  twenty  rows,  with  sixteen  in  each  row ;  the  area  of  each 
part  will  be  -J-x|  =  J;  and  the  perpendicular  distance 
from  the  neutral  line  to  the  centres  of  these  320  parts 
will  be  : 

In  the  first  row,     J 

"      second  "        f 

"      third       "        J 

"      fourth    "         I 

and  so  on,  each  distance  being  a  fraction  having  4  for  a  de- 
nominator, and  for  a  numerator  one  of  the  arithmetical  series 
of  the  odd  numbers  i,  3,  5,  7,  9,  11,  etc.,  to  39.  The 
moment  of  inertia  will  be  the  sum  of  the  products,  as  follows : 

In  the  first  row,  /,  =  i6x£x(£)2 
"  second  "  /,=  i6xix(£)2 
"  third  "  /,  =  1 6  x  i  x  (£)2  etc. 

These  are  equal  to : 

In  the  first  row,  7,  =  16  x  ix^  x  i2  =  £x  ia 
"  second  "  I2  =  16  x  \  x  -^  x  f  =  £  x  32 
"  third  u  I3=  i6xix-^x  5'  =  £  x  5'  etc. 

Thus  the  sum  of  all  the  products  will  be  equal  to  a  quarter 
of  the  sum  of  the  squares  of  the  arithmetical  series  of  the 
odd  numbers  i,  3,  5,  7,  9,  11,  etc.,  to  39. 

Selecting  the  squares  of  these  numbers  from  a  table  of 
squares,  we  find  their  sum  to  equal  10,660,  and  then,  as 
above, 

/  =  £  x  10660  =  2665 


318  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

4-61. — Comparison  of  ResuBt§. — We  have  now  the  three 
results,  2640,  2660,  and  2665,  gradually  increasing  as  the 
number  of  parts  into  which  the  sectional  area  is  divided  in- 
creases, and  tending  towards  the  true  amount,  to  which  it  can 
only  arrive  when  the  parts  become  infinitely  small  and  in- 
finite in  number.  To  compute  these  by  the  arithmetical 
method  would  be  impossible,  but  by  the  calculus  it  is  exceed- 
ingly simple  and  direct.  The  formula  for  the  moment  of 
inertia,  as  generally  used,  is  complicated,  but  for  a  rectangu- 
lar section  in  a  horizontal  beam,  subject  to  limited  vertical 
pressure,  is  simple. 

462. — Moment  of  Inertia,  toy  the  Calculu§— Preliminary 
Statement. — Let     A  BCD     in  Fig.  70  represent  the  rectangu- 
lar cross-section  of  a  beam  ;  let     MN    be  the  neutral  line, 
and  the  two  lines  at     EF    be  drawn  parallel  to    MN.     Let 
the  breadth  of  the  section     EF    equal    7, 

JM 

and  the  perpendicular  distance  from  the 
neutral  line  to  the  lower  line  EF  equal 
x.  The  two  parallel  lines  at  EF  may  be 
taken  at  any  distance,  x,  from  the  neu- 
tral line.  This  distance  is  variable ;  x  is 
a  variable  representing  any  and  every  dis- 
tance possible  on  the  line  GH,  from  zero 
to  its  full  length.  It  is  always  the  distance 
from  the  line  MN  to  7,  the  lower  line 
at  EF,  wherever  7  be  taken.  The  ver- 
tical distance  between  the  two  lines  at 
EF  is  termed  dx,  which  means  the  differential  of  x,  or 
the  difference  in  the  length  of  x  when  slightly  increased 
by  the  movement  of  7  farther  from  MN.  This  augmen- 
tation, dx,  is  taken  infinitesimally  small. 

Now  the  area  of  the  space  between  the  two  lines  at  EF 
will  be  the  product  of  its  length  by  its  height,  or  7  x  dx. 


M- 


MOMENT   OF   INERTIA,  43Y   THE   CALCULUS.  319 

4-63.  —  Moment    of  Inertia,  by  the    Calculus.  —  The    mo- 

ment of  inertia  is  equal  (Art.  457)  to  the  sum  of  the  products 
of  each  particle  of  the  area  of  the  cross-section,  into  the 
square  of  its  distance  from  the  neutral  axis.  In  the  last 
article,  the  expression  ydx  represents  the  area  of  the  in- 
finitesimally  small  space  at  the  lines  EF,  Fig.  70.  The  dis- 
tance from  this  small  area  to  the  neutral  axis  is  x,  and  the 
square  of  the  distance  is  x*  ;  therefore  x*ydx  equals  the 
area  into  the  square  of  its  distance,  equals  the  moment  of 
inertia  of  the  small  area  ydx\  or,  the  differential  of  the 
moment  of  the  area  of  the  whole  figure  ABMN.  This 
differential  is  expressed  thus, 

dl=x*ydx  (203.) 

This  expression  represents  the  moment  of  only  one  of  the 
infinitesimal  parts  into  which  the  area  ABMN  is  supposed 
to  be  divided.  To  obtain  the  moment  of  the  whole  area,  it 
is  requisite  to  add  together  the  moments  of  all  the  infinitesi- 
mal parts  ;  or,  to  obtain  from  the  differential  (form.  203.)  its 
integral.  The  rule  for  this  is,*  "  Add  one  to  the  index  of  the 
variable,  and  divide  by  the  index  thus  increased  and  by  the 
differential  of  the  variable."  Applying  this  rule  to  formula 
(203.)  it  becomes 


This  is  in  its  general  form.  To  make  it  definite,  we  have 
y  =  b,  the  breadth  ;  and  -r,  at  its  maximum,  equals  %d, 
half  the  depth.  These  values  substituted  for  y  and  x> 
we  have 


(204) 


*  Ritchie,  Dif.  and  Integ.  Calculus,  p.  21. 


320  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

This  result  is  the  moment  of  inertia  for  the  upper  half  of  the 
section  of  the  beam,  and  represents  the  resistance  to  com- 
pression. The  resistance  to  tension  in  the  lower  half  of  the 
beam  is  (under  the  circumstances  of  the  case  we  are  consider- 
ing) an  equal  amount  ;  hence  for  the  two  we  have* 


(205.) 


464-.  —  Application  and  Comparison.  —  This  formula  gives 
the  value  of  the  moment  of  inertia  for  the  whole  section;  for 
the  two  parts,  one  above  and  the  other  below  the  neutral 
line.  To  obtain  the  value  of  the  part  above  the  line,  for  com- 
parison with  the  results  obtained  in  Arts.  4-58  to  460,  we 
take  formula  (204.) 


in  which  b  is  the  breadth  and  d  the  depth  of  the  beam. 
The  section  of  beam  given  in  Art.  4-58,  Fig.  69,  was  proposed 
to  be  8  inches  broad  and  20  inches  high,  or  AB  =  b  =  8 
and  AD  —  d—  20.  With  these  figures  in  the  formula,  we 
have 

/  =  ^  x  8  x  203  =  2666f 

This  is  the  exact  amount.  In  the  three  trials  of  Arts.  4-58 
to  4-60,  we  had  the  approximate  values  2640,  2660  and 
2665.  In  the  last  trial,  in  which  the  parts  were  small  and 
numerous,  the  result  was  a  close  approximation. 


*  Moseley,  Am.  Ed.  by  Mahan,  Art.  362. 


MOMENT   OF   INERTIA — GRAPHICAL   REPRESENTATION.    321 

465. — Moment    of  Inertia    Graphically    Represented. — 

The  two  processes,  arithmetical  and  by  the  calculus, 
are  graphically  represented  in  Y 
Fig.  71,  in  which  the  area  of  the 
figure  contained  within  the  straight 
lines  OB  and  AB  and  the  curved 
line  OA,  is  the  correct  area  by 
the  calculus,  to  which  the  sum  of 
the  squares  of  the  arithmetical 
progression  I,  3,  5,  7,  9  and  u 
closely  approximates.  Here  x 
and  y,  indicating  the  distances 
along  the  axes  OX  and  OY,  are 
co-ordinates  to  points  in  the  curve, 
as  Aj  C,  D,  E,  etc.,  these  points 
being  midway  in  the  difference  be- 
tween the  sides  of  each  two  con- 
tiguous squares.  The  values  of  y 
for  these  points  are  2,  4,  6,  8, 
10  and  12  ;  a  difference  between  FIG.  71. 

each  two  consecutive  values  equal  to  2.  The  consecutive 
ordinates  x  are  i,  4,  9,  16,  25  and  36;  or  i\  2\  3", 
42,  52  and  62. 

Comparing  these  values  ol    y    and   x  in  each  pair,  we 
have 


In  the  first     pair,      y 

"     "  second  " 

"     "  third      " 

"     "  fourth    "' 

"     "  fifth     ,  " 

"     "  sixth       " 


y  = 

2 

and      x  — 

I 

=  i" 

y  - 

4 

X  — 

4 

=  22 

y  = 

6 

"            X  ~~~ 

9 

=  3' 

y  = 

8 

X  — 

16 

=  4' 

y  ~~~ 

10 

"            X  = 

25 

=  5° 

y  — 

12 

x  — 

36 

=  6' 

322  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

From  this,  the  relation  between    y   and   x   is  readily  seen  to 
be  represented  by  the  following  expressions : 

(D-  -  =  "v 

f  =  4*  (206.) 


4-66.  —  Parabolic  Curve  — Area  of  Figure. — The  ex- 
pression just  obtained  is  the  equation  to  the  curve,  and 
this  curve  is  a  parabola,  with  /  =  2,  or  y*  —  2px.*  By 
formula  (206.)  any  number  of  points  in  the  curve  may  be 
found,  and  the  curve  itself  drawn  through  them.  Also,  by 
it  and  by  the  rules  of  the  calculus,  the  area  of  the  figure 
inclosed  between  the  curved  line  and  the  two  lines  AB  and 
BO  may  be  found.  To  this  end,  let  the  narrow  space  in- 
cluded between  the  two  lines  GH,  drawn  perpendicular  to 
OB  from  H  to  G  (a  point  in  the  curve),  be  a  small  por- 
tion of  the  area  of  the  whole  figure  ;  dx,  the  distance  be- 
tween the  two  lines,  being  exceedingly  small.  The  area  of 
this  narrow  space  will  be  the  product  of  its  length  by  its 
breadth,  or  y  x  dx.  The  differential  of  formula  (206.),  the 
equation  to  the  curve,f  is 

2ydy  =  Afdx 
\ydy  —  dx 

Multiplying  both  sides  by   y    gives 

=  ydx 


*  Robinson's  Conic  Sections  and  Analytical  Geometry,  1863,  p.  50. 
f  Ritchie's  Dif.  and  Integ.  Calculus,  p.  20. 


MOMENT   OF   INERTIA  —  PARABOLIC    CURVE.  323 

which  equals  the   differential   of  the  area  as  above  shown. 
The  integral  of  this  value  of   ydx    is,  by  the  rule  (Art.  4-63), 


\ffdy  = 


or  the  area 


This  is  the  area  of  the  figure  bounded   by  the  curved  line 
OA    and  the  straight  lines   AB   and   BO. 


467.  —  Example.  —  The  example  given  in  Art.  460  may 
be  taken  to  show  an  application  of  the  last  formula.  The 
number  of  horizontal  rows  of  parts  into  which  the  area  is 
there  divided  is  20,  and  the  last  number  of  the  arithmetical 
series  is  39.  By  an  examination  of  Fig.  71,  it  will  be  seen 
that  AB,  the  base  of  the  figure,  is  equal  to  the  side  ol  the 
last  square  plus  unity.  Therefore,  39+1  =40  is  the  base  of 
the  area  proposed  in  Art.  460,  or.  y  =  40.  From  the  dis- 
cussion in  that  article,  it  appears  that  the  small  squares  con- 
sidered are  each  J  of  unity  in  area,  from  which  the  area 
of  the  figure  in  that  case  is  found  to  be  one  quarter  of  the 
sum  of  the  squares  of  the  arithmetical  series  ;  or,  by  formula 
(207.}, 

A  =  i  x  i/  = 


To  apply  this  result  to  the  present  case,  where    y  =  40,    we 
have 

A  =  -fa  x  4O3  =  2 
the  same  result  as  in  Art.  464. 


324  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

468.  —  Moment  of  Inertia—  General  Rule.  —  That  formula 
(207  '.)  may  be  general  in  its  application,  we  need  to  find  a 
proper  coefficient. 

Let  the  beam,  instead  of  being  8  inches  wide,  as  in  Fig. 
69,  be  only  one  inch  wide,  and  let  the  portion  above  the 
neutral  line  be  divided  by  horizontal  lines  into  any  number 
of  equal  parts.  Put  n  for  the  number  of  parts,  and  /  for 
the  thickness  of  each  part.  The  area  of  each  part  will  be 
I  x  t  =  t  inches,  and  the  several  distances  from  the  neutral 
line  to  the  centre  of  each  part  will  be,  respectively,  -J/,  f/, 

^  2^  _     T 

|/,   |^,    etc.,  to  the  last,  which  will  be   "•-  -  1. 

Now,  the  moment  of  inertia  of  each  part  being  its  area 
into  the  square  of  the  distance  to  its  centre  of  gravity,  there- 
fore the  several  moments  will  be  as  follows  : 

In  the  first  piece,          t  (  i  x  -J  ==  i2  x  \f 

second  "  /  (3  x  ^)  =  3*  x  i'3 

(       t  \2 
third       "  /^5  x  -J  =  52  x^t3 

f        t  \2 
fourth    "  /  ^7  x  -J  =  f  x  J/3 

last         "       t  ((2n—  i)^J  =  (2n—  i)2  x  \f 
The  sum  of  these  will  be 

5=  i/3[i2+  32+  52+72+  ......  (2n—  i)2]     (208.) 

But  the  sum  of  the  series  I2+  3"  +  52  +  etc.,  is  the  area  of 
the  parabolic  figure  (Fig.  71),  ancl  has  been  found  to  be  equal 
to  \f  (form.  207.) 


"     " 


"     " 


MOMENT   OF   INERTIA  —  RULE.  325 

Now  y,  when  at  its  maximum,  coincides  with  the  base 
AB  of  Fig.  71,  and  is  equal  to  the  side  of  the  last  square 
plus  unity.  As  above,  the  side  of  the  last  square  is  2n  —  I, 
from  which  y  —  2n,  and 


and  therefore  formula  (208.)  becomes 


5  =  \fvt  (209.) 

which  is   a   rule  for  ascertaining   correctly  the  moment  of 
inertia  for  a  beam  one  inch  broad. 


469.  —  Application.  —  To  show  the  application  of  the 
above,  take  the  example  of  Art.  458,  where  the  number  of 
slices  is  5  and  the  thickness  is  2,  and  we  have,  by  the  use 
of  formula  (209.  \ 


The  formula  gives  the  result  for  a  beam  one  inch  broad. 
The  beam  in  Art.  458  is  8  inches  broad.  Therefore,  for 
the  full  amount  we  have 

8  x  3334  =  2666f 

Again,  take  the  example  of  Art.  460,  where   t  =  %   and 
n  =  20,    and  we  find  as  the  result 


and  8  x  333^-  =  2666f 

Thus  in  both  cases  we  have  the  same  result  as  that  obtained 
directly  by  the  calculus.  If  b,  for  the  breadth,  be  added 
to  formula  (209.)  we  shall  have  the  complete  rule,  thus  : 

/  =      fif 


326 


ROLLED-IRON    BEAMS. 


CHAP.    XIX. 


and    since     ///    equals    the    height    above   the    neutral   line, 
equals    —  ,    the  half  of  the  depth  of  the  beam, 

t3n3  =  (tn)3  &  (%dj  m  \ds 

and  this  value  of    tsn3    substituted  for  it  in  the  above  equa- 
tion, gives 


This  is  for  one  half  the  beam.     For  the  whole  beam  we  have 
twice  this  amount,  or 


the  same  as  found  directly  by  the  calculus  in  formula  (205.} 


470.  —  Rolled-Iron    Beam  —  Moment   of   Inertia  —  Top 
Flange. — An  expression  for  the  moment  of  inertia  appropri- 
ate to  rolled-iron  beams  of  the    I    form  of  section  (Fig.  68) 
A  B  may  be  obtained  directly  from  the  for- 

mula (205.)  for  the  rectangular  section. 
In  Fig.  72,  showing  the  cross-section 
required,  b  equals  the  breadth  of  the 
beam,  or  the  width  of  the  top  and  bot- 
tom flanges,  and  t  equals  the  width 
or  thickness  of  the  web  ;  b  minus  t 
equals  b/y  d  equals  the  entire  height 
of  the  section,  and  df  the  height  be- 
tween the  flanges.  MN  is  the  neu- 
FlG-  72.  tral  line  drawn  at  half  height. 

By  formulas  (203.)  and  (204.)  the  moment  of  inertia  for 
the  part  above  the  neutral  axis  is 


M 


T  


MOMENT   OF  INERTIA   FOR  FLANGE   AND   WEB.  327 

If  this  be  applied  so  that  x  =  \d,  the  result  (£%bd3)y  as  in 
(204*),  is  the  moment  for  the  rectangle  ABMN.  Again,  if 
it  be  applied  with  x  =  \dt,  the  result  (^bdf)  will  be  the 
moment  for  the  rectangle  EFMN.  Now,  if  the  latter  re- 
sult be  subtracted  from  the  former,  the  remainder  will  be 
the  moment  for  the  area  ABEF,  the  upper  flange,  or 


4-71.—  Rolled-Iron     Beam—  Moment    of    Inertia—  Web.— 

Formula  (210.)  is  the  moment  of  inertia  for  the  top  flange. 
The  moment  of  inertia  for  the  upper  half  of  the  web  is  that 
due  to  a  rectangle  having  for  its  breadth  y  =  t,  and  for  its 
height  x  =  \dt,  and  by  Art.  463, 


and  since    /  =  b  —  bt,    therefore 


4-72.—  Rolled-Iron  Beam— moment  of  Inertia— Flange 
and  Web. — Formula  (211.)  is  the  moment  of  inertia  for  the 
upper  half  of  the  web.  Added  to  formula  (210.),  the  sum, 
representing  the  moment  of  inertia  for  all  of  the  beam  above 
the  neutral  line,  will  be 


328  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

4-73.  —  Rolled-Iron  Beam  —  Moment  of  Inertia  —  Whole 
Section.  —  Formula  (212.)  is  the  moment  of  inertia  for  that 
half  of  the  rolled-iron  beam  which  is  located  above  the  neu- 
tral line.  The  moment  for  the  portion  below  the  line  will 
be  equal  in  amount  ;  and  therefore,  for  the  moment  of  the 
entire  section,  we  have  twice  the  amount  of  formula  (212.)  or 

/  =  TV(WW,//)  (213.) 

474.—  Rolled-Iron  Beam—  Moment  of  Inertia—  Compari- 
son with  other  Formulas.  —  Formula  (213.)  is  the  same  as 
that  given  by  Professor  Rankine*  and  others,  and  is  in  gen- 
eral use.  Canon  Moseleyf  gives  an  expression  which  is 
complicated.  Mr.  Edwin  Clark,  in  his  valuable  work  on  the 
Britannia  and  Conway  Tubular  Bridges,  Vol.  I.,  p.  247,  gives 
the  formula 


in  which  d2  is  the  distance  between  the  centres  of  gravity 
of  the  top  and  bottom  flanges,  a  is  the  area  of  the  top  or 
bottom  flange,  and  a/  is  the  area  of  the  web.  This  is 
more  simple  than  the  common  formula  (213.),  but  is  not 
exact.  It  is  only  an  approximation.  Its  relation  to  the  true 
formula  will  now  be  shown. 

From  formula  (213.)  we  have,  multiplying  by     12, 


Of  these  symbols  we  have  (Fig.  72,  putting    h  =  AE  ), 

d  =  d2  +  h,      bt  =  b—t     and     dt  =  d2—h 
By  substitution,  we  now  have 

1  2/  =  b(d2  +  //)'  -  (b-t}df 


*  Rankine's  Applied  Mechanics,  pp.  316  and  317. 

f  Moseley's  Mech.  of  Eng.,  Am.  Ed.  by  Mahan,  Art.  504. 


MOMENT   OF   INERTIA  —  FORMULAS    COMPARED.  329 

and  since  (b—t)df  —  bd?—td?  =  bd?—atd*  (putting  at  for 
the  area  of  the  web);  and  since  dj  —  d2—h,  therefore  we 
have 


=  b(d2  4-  h)3  -  \b(d2-h)3-atd? 
1  2/  =  b(d,  4-  //)*  -  b(ds—h)3  +  atdf 


Then  we  have 

(d,  +  /i)3  =  d/  + 


(d2  +  h)3-(d2-/i}3  =       o  +  &////  +  o  4-  2k3 
Substituting  these  in  the  above,  we  have 

I2/  =  b(6dfh  4-  2k3)  +  atdf 
The  area  of  the  top  flange  equals    bh  =  a,    therefore 

1  2/  =  6a(d;+  \h2}  +  atd?  or, 

/  =  &\6a(d;+  i//)  +  *,<//]  (215.) 

In  Mr.  Clark's  formula,  (2  14-),  we  have 

+  «y)  or, 

+  a.df) 


Comparing  this  with  the  reduction  of  the  common  formula 
as  just  found  [form.  (215.)'],  the  difference  is  readily  seen 
to  be,  that  while  in  the  one  the  quantities  a  and  at  are 
each  multiplied  by  the  factor  df,  in  the  other  the  factor  for 
a  is  (X/+i//)  and  that  for  at  is  df. 


330  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

475.—  Rolled-Iron  Beam—  moment  of  Inertia—  Compari- 
son of  Itctult*.  —  To  show,  by  an  application,  the  difference 
in  the  results  obtained  by  the  two  formulas  (214-}  and  (215.), 
let  it  be  required  to  find  the  moment  of  inertia  for  a  rolled- 
iron  beam  12  inches  high  and  4  inches  broad,  and  in 
which  the  top  and  bottom  flanges  are  one  inch  thick,  and 
the  web  one  half  inch  thick.  Here  we  have  d—  12,  dt  —  10, 
<4=n,  /  =  J,  £=4,  ^  =  3i,  #  =  4x1=4,  a/ 
and  h  =  i  ;  and  by  formula  (214)  we  have 

/=  .^x  1  12  x  (6x4  +  5)  =  292^5- 
The  value  by  formula  (215.)  is 


The  value  by  the  common  formula,  (213.),  is 

1  =  AK4  x  I23)~(3  •  5  x  io3)]  =  284!- 

Thus  we  have  by  either  of  the  two  formulas  (213..)  or  (215.) 
the  exact  value,  /=  284^,  while  by  formula  (214)  the 
value  obtained  is  /=  292^. 


476  • — Rolled-Iron  Beam — moment  of  Inertia — Remarks. 

—When,  in  a  rolled-iron  beam,  the  top  and  bottom  flanges 
are  comparatively  thin,  the  difference  between  dt  and  da 
will  be  small,  and  in  consequence  the  value  of  7  as  derived 
by  formula  (214.)  will  differ  but  little  from  the  truth.  This 
formula,  therefore,  for  such  cases,  is  a  near  approximation, 
and  for  some  purposes  may  be  useful ;  but  formula  (215.), 
and  that  from  which  it  is  derived,  (213.),  are  exact  in  their 
results,  and  should  be  used  in  preference  to  formula  (2 14-) 
in  all  important  cases. 


LOAD   AT   MIDDLE  —  RULES.  331 

477.  —  Reduction   of    Formula—  Load   at    Middle.  —  The 

expression   (213.),   then,   is   that   which   is   proper   for    the 
moment  of  inertia  for  rolled-iron  beams  —  namely  : 


In  Art.  303,  formula  (115.),  we  have 

wr 


"   16 

This  is  for  a  beam  supported  at  each  end,  with  the  load  in 
pounds  at  the  middle,  the  length  in  feet  and  the  other 
dimensions  in  inches.  F  is  a  constant,  which,  from  an 
average  of  experiments  (Art.  701)  upon  rolled-iron  beams, 
has  been  ascertained  to  be  62,000.  The  value  of  /,  the 
moment  of  inertia,  has  been  computed,  for  many  of  the  sizes 
of  beams  in  use,  by  formula  (213.),  and  will  be  found  in 
Table  XVII. 

We  have,  therefore,  from  (115.) 

Wls 

12  X  62000  =-vr- 
16 

744000  =  ~  (216) 

478.— Rules-Values  of  JF,  I,  6  and  JT.— Rule  (216)  is 
for  a  load  at  the  middle  of  a  rolled-iron  beam.  The  values 
of  the  several  symbols  in  (216)  may  be  had  by  transpositions, 
as  follows : 


The  weight,  W=  (217.) 


length,  7  =  ^44^. 

Wl8  t 

deflection,  6  =  -,  (219.) 

744000/ 

Wla 
moment  of  inertia,  /=  - 


744000<5 


332  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

479.— Example— Weight.— Formula  (217.)  is  a  rule  by 
which  to  find  the  weight  in  pounds  which  may  be  carried  at 
the  middle  of  a  rolled-iron  beam,  with  a  given  deflection. 
As  an  example :  What  weight  may  be  carried  at  the  middle 
of  a  9  inch  90  pound  beam,  20  feet  long  between  bear- 
ings, with  a  deflection  of  one  inch  ? 

Here  we  have  6  =  i,  /  =  20  and  (from  Table  XVII.) 
/=  109-117  ;  and,  by  the  formula, 

TT_      744000  x  log-  H7  x  i 

W=/-  -^-  -=10147-881 

or  the  weight  to  be  carried  equals    10,148   pounds,  or  say    5 
net  tons. 


480. — Example— Length. — Formula  (218.)  is  a  rule  by 
which  to  find  the  length  at  which  a  beam  may  be  used  when 
required  to  carry  at  the  middle  a  given  load,  with  a  given 
deflection.  For  example  :  To  what  length  may  a  Buffalo 
6  inch  50  pound  rolled-iron  beam  be  used,  when  required 
to  carry  5000  pounds  at  the  middle,  with  a  deflection  of 
-^  of  an  inch  ? 

Here  7=29-074  (from  Table  XVII. ),  6  =  0-3  and 
W  —  5000  ;  and,  by  the  iormula, 


/=  //7440QQX  29-074x0.3  =  IQ 

5OOO 

or  the  length  may  be    10   feet    1 1    inches. 


481. — Example— Deflection. — Formula  (219.)  is  a  rule 
for  finding  the  deflection  in  a  rolled-iron  beam,  when  carry- 
ing at  the  middle  a  given  load.  As  an  example  :  What  de- 
flection will  be  caused  in  a  Phoenix  9  inch  70  pound  beam 
20  feet  long,  by  a  load  of  7500  pounds  at  the  middle  ? 


LOAD   AT   ANY    POINT  —  GENERAL   RULE.  333 

Here      ^=7500,      /  =  20      and    (from    Table    XVII.) 
/  =  92-207  ;  and,  by  the  formula, 

75OO  X  2O8 

=087461 


744000  x  92  •  207 
or  the  deflection  will  be    -J    of  an  inch. 


482. — Example— moment  of  Inertia. — In  formula 
we  have  a  rule  by  which  to  ascertain  the  moment  of  inertia 
of  a  rolled-iron  beam,  laid  on  two  supports,  and  carrying  a 
load  at  the  middle.  To  exemplify  the  rule  :  Which  of  the 
beams  in  Table  XVII.  would  be  proper  to  carry  10.000 
pounds  at  the  middle,  with  a  deflection  of  one  inch  ;  the 
length  between  the  bearings  being  twenty  feet  ? 

Here    W  —  10000,   /  =  20    and    6  =  i,    and  by  the  for- 
mula, 

i oooo x  2o3 

/  —  -  -  =  107-527 

744000  x  i 

or  the  required  moment  of  inertia  is  107-527.  The  nearest 
amount  to  this  in  Table  XVII.  is  107-793,  pertaining  to  the 
Phoenix  9  inch  84  pound  beam.  This  beam,  therefore, 
would  be  the  one  required. 


483.— Load  at  Any  Point— General  Rule. — The  rules 
just  given  are  for  cases  where  the  loads  are  at  the  middle. 
Rules  for  loads  at  any  other  place  in  the  length  will  now  be 
developed. 

Formula  (%3*\  is 


334  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

If     bd*     be    multiplied    by     -^    its    value    will    not    be 
changed,  and  there  will  result 
\2bd3       i2 


and  formula   &f.   becomes 


Bv  formula  (154.\  in  ^4r/.  376, 

Bl 


and  as     rl  =  6,     or     r  =  -j,     therefore 


a~ 


d  -  Fdd 
Fdj 


For     a    in  the  above,  substituting  this  value,  we  have 


mn        _ 


- 
a 

=  12/73 


484.  —  Load  at  Any  Point  on  Rolled-Iron  Ream*.  —  The 

moment  of  inertia,  /,  in  formula  (221.)  is  [form.  (205.)'], 
I  —  -£?bd3  for  a  rectangular  beam.  For  a  tube,  or  for  a 
beam  of  the  I  form,  it  is,  by  formula  (213.), 


If  in  (221)  we  substitute  for    /   this  value  of  it,  we  have 
iWlmn  =  Fd(bd3-b{d?)  (222.) 


LOAD   AT  ANY   POINT.  335 

This  is  a  rule  for  rolled-iron  beams  supported  at  each  end 
and  carrying  a  load  at  any  point  in  the  length,  with  a  given 
deflection  ;  and  in  which  W  is  the  weight  in  pounds,  m 
and  n  the  distances  from  the  load  to  the  two  supports, 
and  m  plus  n  equals  /  equals  the  length  ;  m,  n  and  / 
all  being  in  feet  ;  6  is  the  deflection,  b  and  d  are  the 
breadth  and  depth  of  the  beam,  bt  and  dt  the  breadth  and 
depth  of  the  part  which  is  wanting  of  the  solid  bd  (Art. 
470)  ;  <?,  b,  d,  bt  and  dt  all  being  in  inches  ;  and  F  is 
the  constant  for  rolled  iron  (Table  XX.). 

485.  —  Load  at  Any  Point  on  Rolled-Iron  Beams  of 
Table  XVII.  —  The  value  of  F  is  62,000.  If  it  be  substituted 
for  F  in  (221)  we  shall  have 

4  Wlmn  —  1  2  x  62OOO/d 
^Wlmn  =  744000!  d 
Wlmn  =  i86ooo/<? 

i 

W  = 


. 
Imn 

which  is  a  rule  for  ascertaining  the  weight  which  may  be 
carried,  with  a  given  deflection,  at  any  point  in  the  length 
of  any  of  the  rolled-iron  beams  of  Table  XVII. 

486.  —  Example.—  What  weight  may  be  carried  on  a 
Paterson  12^  inch  125  pound  rolled-iron  beam,  25  feet 
long  between  bearings,  at  10  feet  from  one  of  the  bearings, 
with  a  deflection  of  1-5  inches  ? 

Here  we  have  6  =  1-5,  m  —  10,  n  =  15,  /  =  25  and 
/  =  292-05  (from  Table  XVII.)  ;  and  hence 

186000x292.05x11 

25  x  lox  15 
or  the  weight  allowable  is,  say    21,730    pounds. 


336  ROLLED-IRON    BEAMS.  CHAP.    XIX. 

487. — Load  at  End  of  Rolled-Iron  Lever. — In  formula 
(113.)  we  have 

Wl3 
12F=~I6-  or> 

Wl'  =  I2FI6 

This  expression  is  for  a  beam  supported  at  each  end  and 
loaded  at  the  middle.  In  a  lever  the  strains  will  be  the  same 
when  the  weight  and  length  are  each  just  one  half  those  in  a 
beam  supported  at  each  end.  Hence  if  for  W  we  take 
2P,  and  for  /  take  2n,  P  being  the  weight  at  the  end 
of  a  lever  and  n  the  length  of  the  lever,  we  shall  have,  by 
substitution  in  the  above, 

2Px  2H3  =   \2FId 

i6Pn*=  \2FId 

i6Pn 3  =  Fd  (bd3-btdf)  (224.) 

and  Pn3  =  %Fl6  (225.) 


and  further,  since     F=  62000    (Table  XX.),  therefore 

Pns  =  465001$ 
46  5  oo IS 


P=        n~ 


which  is  a  rule  for  ascertaining  the  weight  which  may  be 
supported  at  the  free  end  of  a  lever,  with  a  given  deflection, 
the  lever  being  made  of  any  one  of  the  rolled-iron  beams  of 
Table  XVII. 


488. — Example. — Let  it  be  required  to  show  the  weight 
which  may  be  sustained  at  the  free  end  of  a  Trenton  15^- 
inch  150  pound  rolled-iron  beam,  firmly  imbedded  in  a  wall, 
and  projecting  therefrom  20  feet;  the  deflection  not  to 
exceed  2  inches. 


LOAD    UNIFORMLY    DISTRIBUTED.  337 

Here    7=528-223    (Table  XVIL),     <S  =  2    and    n  =  2O', 
and  by  formula  (226.) 


or  the  weight  which  may  be  carried  is    6140   pounds. 


489. — Uniformly     Distributed     Load     on     Rolled-Iron 
Beam. — By  formula  (115.)  we  have 


This  is  for  a  load  at  the  middle  of  a  beam.  Let  U  rep- 
resent an  equally  distributed  load;  then  %U  will  have  an 
effect  upon  the  beam  equal  to  the  concentrated  load  W, 
(Art.  340),  and  hence,  substituting  this  value, 

f£//3=  12FI6 

\Ul3  =  F6  (bd3-btd^j  (227.) 

By  Table  XX.     F  =  62000,     and  the  formula  reduces  to 

Ul3  —  1190400/6 
u=^°ol«  ^ 

which  is  a  rule  for  ascertaining  the  amount  of  weight,  equally 
distributed,  which,  with  a  given  deflection,  may  be  borne 
upon  any  of  the  rolled-iron  beams  of  Table  XVIL 


490. — Example. — What  weight,  uniformly  distributed, 
may  be  sustained  upon  a  Buffalo  10^  inch  105  pound 
rolled-iron  beam;  25  feet  long  between  bearings,  with  a 
deflection  of  f  of  an  inch  ? 


338  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

Here    7=175-645   (Table  XVII.),   rf  =  o-75   and    7=25; 
and  therefore,  by  (228.), 

u=  1190400  xj 7^645 ^0.75 
25 

or  the  weight  uniformly  distributed  is     10,036     pounds. 


491.  —  Uniformly  Distributed  Load  on  Rolled-Iron 
Lever.  —  A  rule  for  a  lever  loaded  at  the  free  end  is  given  in 
formula  (225.), 

Pn3  = 


When  a  load  concentrated  at  the  free  end  of  a  lever  is 
equal  to  f  of  a  load  uniformly  distributed  over  the  length 
of  the  lever,  the  effects  are  equal.  (Art.  347.)* 

If  U  equals  the  load  equally  distributed,  and  P  the 
load  concentrated  at  the  free  end,  then  f  U  —  P,  and  sub- 
stituting this  value  for  P  in  formula  (225.)  gives 


=  \2FI6 

6  Un3  =  F6  (bds-btdf)  (229.) 

Putting  for     F    its  value     62,000,     and  reducing,  we  have 

Uns  —  1  24.000/6 

124.000/6 

~~" 


which  is  a  rule  for  ascertaining  the  load,  uniformly  distrib- 
uted, which  may  be  sustained  upon  any  of  the  rolled-iron 
beams  of  Table  XVII.  ,  with  a  given  deflection,  when  used 
as  a  lever. 

*  Rankine,  Applied  Mechanics,  p.  329. 


LOAD   ON  A  FLOOR — ITS   COMPONENTS.  339 

492. — Example. — What  weight,  uniformly  distributed, 
may  be  sustained  upon  a  Trenton  6  inch  40  pound 
rolled-iron  beam,  used  as  a  lever,  and  projecting  10  feet 
from  a  wall  in  which  it  is  firmly  imbedded;  the  deflection 
not  exceeding  f  of  an  inch  ? 

Here     /  =  23  •  761 ,     6  =  f     and     n  =  10  ;  and  by  (230.) 


ioj 
or  the  weight  will  be     1965     pounds. 

493. — Component*  of  Load  on  Floor. — When  rolled- 
iron  beams  are  used  as  floor  beams,  they  have  to  sustain  a 
compouncj  load.  This  load  may  be  considered  as  composed 
of  three  parts,  namely  : 

First :  The  superincumbent  load,  or  load  proper; 

Second :  The  weights  of  the  materials  within  the  spaces 
between  the  beams,  and  of  the  covering  ;  and, 

Third :  The  weight  of  the  beams  themselves. 

494. — The  Superincumbent  Load. — This  will  be  in  pro- 
portion to  the  use  to  which  the  floor  is  to  be  subjected.  If 
for  the  storage  of  merchandise,  the  weight  will  vary  accord- 
ing to  the  weight  of  the  particular  merchandise  intended  to 
be  stored.  Warehouses  are  sometimes  loaded  heavily,  and 
for  these  each  case  needs  special  computation.  For  general 
purposes,  such  as  our  first-class  stores  are  intended  for,  the 
load  may  be  taken  at  250  pounds  per  superficial  foot  (Art. 
368).  A  portion  of  the  floor  may  in  some  cases  be  loaded 
heavier  than  this,  but  as  there  is  always  a  considerable  part 
kept  free  for  passage  ways,  250  pounds  per  foot  will  in 
general  be  found  ample  to  cover  the  heavier  loads  on  floors 
of  this  class. 


340  ROLLED-IRON    BEAMS.  CHAP.    XIX. 

On  the  floors  of  assembly  rooms,  banks,  insurance  offices, 
dwellings,  and  of  all  buildings  in  which  the  floors  are  likely 
to  be  covered  with  people,  the  weight  may  be  taken  at  66, 
or  say  70  pounds  per  foot  ;  66  pounds  being  the  weight 
of  a  crowd  of  people  (Art.  114). 


495. — The  Materials  of  Construction— Their   Weight.— 

These  (not  including  the  iron  beam)  will  differ  in  accordance 
with  the  plan  of  construction.  As  usually  made,  with  brick 
arches,  concrete  filling,  and  wooden  floor  laid  on  strips  bed- 
ded in  the  concrete,  this  weight  will  not  differ  much  from 
70  pounds  per  superficial  foot,  and,  in  general,  it  may  be 
taken  at  this  amount. 


496.  —The  Rolled-Iron  Beam—  Its  Weight.  —  The  differ- 
ence in  the  weight  of  rolled-iron  beams  is  too  great  to  per- 
mit the  use  in  the  rule  of  a  definite  amount,  taken  as  an 
average.  To  represent  this  weight,  therefore,  we  shall  have 
to  make  use  of  a  symbolic  expression. 

Let  y  equal  the  weight  of  the  beam  in  pounds  per  lineal 
yard,  and  c  equal  the  distance  in  feet  between  the  centres 
of  two  adjacent  beams.  Then  \y  will  equal  the  weight  of 
the  beam  per  lineal  foot  ;  and  this  divided  by  c  will  give, 
as  a  quotient, 


equals  the  weight  of  beam  per  superficial  foot  of  the  floor. 


497. — Total  Load  on  Floors. — Putting  together  the 
three  weights,  as  above,  we  have  the  total  weight  per  super- 
ficial foot  as  follows : 


LOAD   PER   SUPERFICIAL   FOOT.  341 

For  the  floors  of  dwellings,  assembly  rooms,  banks,  etc., 

the  superincumbent  load  is  70  pounds  ;   • 

the  brick  arches,  concrete,  etc.,  equal         70       " 

y 

and  the  rolled-iron  beams  equal  — 

These  amount  in  all  to 

/=  HO  +  — 

For  the  floors  of  first-class  stores, 

the  superincumbent  load  is  250  pounds; 

the  brick  arches,  concrete,  etc.,  equal  .        70        " 

y 
and  the  rolled-iron  beams  equal  — 

or,  in  all, 


498.  —  Floor  Beam  §—  Distance  from  Centres.  —  In  formula 
>.)  U  stands  for  the  weight  uniformly  distributed  over 
the  length  of  the  beam.  When  /  is  taken  to  represent  the 
total  load  in  pounds  per  superficial  foot  of  the  floor,  c  the 
distance  apart  in  feet  between  the  centres  of  two  adjacent 
beams,  and  /  the  length  of  the  beam  in  feet,  then 


Substituting    for     U   in    formula   (228.}   its   value    as   here 
shown,  we  have 


342  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

When  r  represents  the  rate  of  deflection  per  foot  lineal 
of  the  beam,  we  have  rl  —  d,  equals  the  whole  deflection. 
Substituting  for  6  in  formula  (234-)  this  equivalent  value 
we  have 


J€*  —          I3 


Again  ;  for    f    substituting  its  value  as  in  (232^)t  we  have 


/ 
( 


?- 
140  +       ,  =  --- 


which  is  a  rule  for  ascertaining  the  distance  apart  from  cen- 
tres between  rolled-iron  beams,  in  the  floors  of  assembly 
rooms,  banks,  etc.,  with  a  given  rate  of  deflection. 


4-99. — Example. — It  is  required  to  show  at  what  dis- 
tance from  centres,  Paterson  io|  inch  105  pound  rolled- 
iron  beams,  25  feet  long,  should  be  placed  in  the  floors  of 
a  bank,  in  which  the  rate  of  deflection  is  fixed  at  0-035  °f 
an  inch. 

Here  we  have  7=191.04  (Table  XVII.),  r  ~  0-035, 
/=  25  and  y  =  105  ;  and  by  (236.) 

85024x191.04x0-035       105 

'  =  -          --          --       =   ' 


DISTANCE   BETWEEN   CENTRES,    IN   DWELLINGS.  343 

or   the   distance   from    centres   should   be,   say    3    feet     4! 
inches. 


5 CO.— Floor  Beams— Distance  from  Centre*— Dwellings 

etc. — If  the  rate  of  deflection  be  fixed,  and  at    0-03    (Art. 
314),  then  formula  (236.),  so  modified,  becomes 


_ 

420 


which  is  a  rule  for  ascertaining  the  distance  apart  from  cen- 
tres of  rolled-iron  beams,  in  the  floors  of  assembly  rooms, 
banks,  etc.,  with  a  rate  of  deflection  fixed  at  0-03  of  an 
inch  per  foot  lineal  of  the  beam. 


501. — Example. — What  distance  apart  from  centres 
should  Buffalo  \2\  inch  125  pound  rolled-iron  beams  25 
feet  long  be  placed,  in  the  floor  of  an  assembly  room  ? 

Here  7=286.019  (Table  XVII.),  7=25  and  7=125; 
and  by  formula  (237.) 

25_l;0frx  286.019  _  125 


or  the  distance  from  centres  should  be    4!    feet,  or    4    feet 
4J    inches. 

The  distances  from  centres  of  various  sizes  of  beams  have 
been  computed  by  formula  (®$7*\  and  the  results  are  re- 
corded in  Table  XVIII. 


344  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

502.—  Floor  Beams—  Distance  from  Centres.  —  If  in  for- 
mula (235.)  we  substitute  for  /  its  value  in  (233.)  we  shall 
have 

ngoAOoIr 


i  90400  Ir 

*-i- 


i  i  goAooIr       y 
320^  ==  -  Zjs-  ---  - 


i i 90400 Ir  y 

y 


320/5  320  x  3 


c  = 


I3  960 


This  is  a  rule  for  ascertaining  the  distance  apart  from  cen- 
tres between  rolled-iron  beams,  in  floors  of  first-class  stores, 
with  a  given  rate  of  deflection. 

503. — Example. — At  what  distance  apart  should  Phoenix 
15  inch  150  pound  beams  25  feet  long  be  placed,  with  a 
rate  of  deflection  of  r  =  0-045  ? 

Here  we  have  7=514-87  (Table  XVII.),  r  =  0-045, 
/=  25  and  y  =  150;  and  in  formula  (238.) 

3720x514-87x0-045        150  _ 

•~~ 


or  the  distance  required  is    5-36    feet,  or    5    feet  4^   inches. 

504.— Floor  Beams— Distance  from  Centres— First-class 
Stores. — If  the  rate  of  deflection  be  fixed,  and  at  0-04  of  an 
inch  (Arts.  3l3f  314  and  368),  then  formula  (238.)  becomes 


DISTANCE   BETWEEN.  CENTRES,  "iX  STORES — FLOOR  ARCHES.  34$ 

» 

which  is  a  rule  for  ascertaining  the  distance  apart  from  cen- 
tres of  rolled-iron  beams,  in  floors  of  first-class  stores,  with  a 
rate  of  deflection  fixed  at  0-04  of  an  inch  per  foot  lineal  of 
the  beam. 


505.  —  Example.  —  At  what  distance  apart  should  Buffalo 
I2j  inch  1  80  pound  rolled-iron  beams  20  feet  long  be 
placed,  in  a  first-class  store? 

Here  7  =  418.945  (Table  XVII.),  I  =  20  and  7=180; 
and,  by  the  above  formula, 


148-8  x  418-945       180 


or  the  distance  from  centres  should  be    7-6    feet,  or    7    feet 
7J    inches  nearly. 

The  distances  from  centres,  as  per  formula  (239!),  have 
been  computed  for  rolled-iron  beams  of  various  sizes,  and 
the  results  are  recorded  in  Table  XIX. 


506, — Floor    Arches— General    €on§iderafion§. — If    the 

spaces  between  the  iron  floor  beams  be  filled  with  brick 
arches  and  concrete,  as  in  Art.  495,  care  is  necessary  that 
these  arches  be  constructed  with  very  hard  whole  brick  of 
good  shape,  be  laid  without  mortar,  in  contact  with  each  other, 
and  that  the  joints  be  all  well  filled  with  best  cement  grout 
and  be  keyed  with  slate.  As  to  dimensions,  the  arch  when 
well  built  need  not  be  over  four  inches  thick  for  spans  of 
seven  or  eight  feet,  except  for  about  a  foot  at  each  spring- 
ing, where  it  should  be  eight  inches  thick,  and  where  care 
should  be  taken  to  form  the  skew-back  quite  solid  and  at 
right  angles  to  the  line  of  pressure. 

In  order  to  economize  the  height  devoted  to  the  floor,  it 


346 


ROLLED-IRON   BEAMS. 


CHAP.    XIX. 


is  desirable  to  make  the  versed  sine  or  rise  of  the  arch  small. 
But  there  is  a  limit,  beyond  which  a  reduction  of  the  rise 
will  cause  so  great  a  strain  that  the  material  of  which  the 
bricks  are  made  will  be  rendered  liable  to  crushing.  Experi- 
ments have  shown  that  this  limit  of  rise  is  not  much  less  than 
ij  inches  per  foot  width  of  the  span,  and  in  practice  it  is 
found  to  be  safe  to  make  the  rise  i£  inches  per  foot. 


507. — Floor  Arclic§— Tie-Rods. — The  lateral  thrust  ex- 
erted by  the  brick  arches  may  be  counteracted  by  tie-rods 
of  iron.  The  arches,  if  made  with  a  small  rise,  will  differ  but 
little  in  form  from  the  parabolic  curve.  Let  Fig.  73  repre- 
sent one  half  of  the  arch  and  tie-rod.  Draw  the  lines  AD 
and  DC  tangent  to  the  points  A  and  C.  Then  AE  =  EB* 


FIG.  73. 

equals  i  of  the  span,  or  i^,  and  DE  —  BC  equals  the 
versed  sine,  or  height  of  the  arch.  If  DE,  by  scale,  be 
equal  to  the  load  upon  the  half  arch  AC,  then  AE  equals 
the  horizontal  strain  ;  or 


DE  :  AE  :: 
v  :       j  :  : 


\  H 
:  H 


(240.) 


in  which     U    is  the  load,  in  pounds,  and     s    is  the  span 
and     v    the  versed  sine,  both  in  feet.     To  resist  this  strain 

*  Tredgold's  Elementary  Principles  of  Carpentry,  Art.  57  and  Fig.  28. 


FLOOR  ARCHES—  TIE-RODS.  347 

the  rod  must  contain  the  requisite  amount  of  metal.  The 
ultimate  tensile  strength  of  wrought-iron  may  be  taken  at  an 
average  of  55,000  pounds  per  inch.  Owing,  however,  to  de- 
fects in  material  and  in  workmanship  (such,  for  instance,  as 
an  oblique  bearing,  which,  by  throwing  the  strain  out  of  the 
axis  and  along  one  side  of  the  rod,  would  materially  increase 
the  destructive  effect  of  the  load),  the  metal  should  be 
trusted  with  not  over  9000  pounds  per  inch.  If  a  rep- 
resent the  area  of  the  tie-rod  in  inches,  then 

90000  =  H 
Substituting  this  value  of    H    in  formula  (240.)  we  have 

9000*  =  g  (*4-Z.) 

For  U  we  may  put  its  equivalent,  which  is  the  load  per 
foot  multiplied  by  the  superficial  area  of  the  floor  sustained 

by  the  rod,  or 

U=cfs 

c  being  the  distance  from  centres  between  the  rods,  and  s 
the  span  of  the  arch,  both  in  feet,  and  f  the  weight  of  the 
brick-work  and  the  superimposed  load,  in  pounds,  or 
70+^.  If  the  arch  be  made  to  rise  \\  inches  per  foot  of 
width,  or  \  of  the  span,  then  8v  =  s,  and  formula  (2Jf.l.) 
becomes 

7Q  +  ? 

a  =  -  --  —cs 
9000 

Putting    q,    the  superimposed  load,  at  seventy  pounds,  we 

have 

140 

a  =  --  cs 
9000 


which  is  a  rule  for  the  area,  in  inches,  of  a  tie-rod  in  a  bank, 
office  building,  or  assembly  room  floor. 


348  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

If    q    be  put  equal  to    250    pounds,  then 

320 

a  =  -     —cs 
9000 

a  =  o  •  03-!  x  cs  (^ 44-) 

which  is  a  rule  for  the  area,  in  inches,  of  a  tie-rod  in  the 
floor  of  a  first-class  store. 

For  general  use,  the  diameter,  rather  than  the  area,  of  the 
tie-rod  is  desirable.     We  have  as  the  area  of  any  rod, 

*,=  .7854^ 
and  therefore       -7854^*  =  o-oi^  x  cs 

and  d—  Vo-oi^cs  (®45.) 

which  is  a  rule  for  banks,  etc.  ;  and 

d  =   t/o -04527^ 
which  is  a  rule  for  first-class  stores. 


508. — Example. — In  a  first-class  store,  with  beams  20 
feet  long,  and  arches  6  feet  span :  What  is  the  required 
diameter  of  tie-rods  ? 

Here  s  =  6,  and  if  there  are  to  be,  say  two  rods  in 
the  length  of  each  arch,  then  c  —  6|,  and  therefore 


d=  Vo- 04527  x  6fx  6=  1-35 

or  the  required  rods  are  to  be    if  inches  diameter. 

Tie-rods  should  be  placed  at  or  near  the  bottom  flange, 
and  so  close  together  that  the  horizontal  strain  between  them 
from  the  thrust  of  the  arch  shall  not  be  greater  than  the 
bottom  flange  of  the  beam  is  capable  of  resisting. 


HEADERS   FOR  DWELLINGS   AND   ASSEMBLY   ROOMS.      349 

509.  —  Headers.  —  In  Art.  381  we  have  the  expression 


a  rule  for  a  header  of  rectangular  section.     We  have  also  in 
formula  (205.) 


or  1  2/  =  bd  3 

Substituting   this     I2/    for     b(d—i)*    in    the    above    equa- 
tion gives 


which  is  a  rule  for  rolled-iron  headers  ;  and  in  which  f  is 
the  load  in  pounds  per  superficial  foot,  n  is  the  length  of 
the  tail  beams  having  one  end  resting  on  the  header,  and  g 
is  the  length  of  the  header ;  n  and  g  both  being  in  feet. 


510. — Hcader§  for  Dwelling*,  etc. — If  in  (2 4?  •)  we  sub- 
stitute for  f  its  value  as  per  formula  (232.),  and  for  F  its 
value  62,000  (Table  XX.),  and  make  r  —  0-03  (Art.  314), 
we  shall  have 


/  — 


38-4x62000x0-03 

140  + 


71424 


which  is  a  rule  for  ascertaining  the  moment  of  inertia  of  a 
rolled-iron  header,  in  a  floor  of  an  assembly  room,  bank,  etc.  ; 
from  which  an  inspection  of  Table  XVII.  will  show  the 
required  header. 


350  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

511.  —  Example.  —  In  the  floors  of  a  bank,  constructed  of 
Buffalo  ioj-  inch  105  pound  beams,  placed  4  feet  from 
centres  :  What  ought  a  header  to  be  which  is  20  feet  long, 
and  which  carries  tail  beams  16  feet  long? 

Here    y  =  105,     c  —  4,     ;/  =  16     and    £•=  20;    and    by 


71424 

or  the  beam  should  be  of  such  size  that  its  moment  of  inertia 
be  not  less  than  266-577.  By  reference  to  Table  XVII.  we 
find  the  beam,  the  moment  of  inertia  of  which  is  next  greater 
than  this,  to  be  the  Pottsville  12  inch  125  pound  beam, 
for  which  7=276-162.  This  may  be  taken  for  the  header, 
although  it  is  stronger  than  needed.  Should  the  depth  be 
objectionable,  we  may  use  two  of  the  Pottsville  io|  inch 
90  pound  beams,  bolted  together;  for  of  this  latter  beam 

I  =  150-763,    and 

2x  150-763  =  301-526 

considerably  more  than  266-577,  tne  result  of  the  computa- 
tion by  formula  (248.).  But  these  two  beams,  although 
nearer  the  required  depth,  yet,  when  taken  together,  weigh 
1  80  pounds  per  yard  ;  while  the  12  inch  beam  weighs  but 
125  pounds.  On  the  score  of  economy,  therefore,  it  is 
preferable  to  use  the  12  inch  beam. 

512.  —  Header§     for     Fir§t-cla§§     Stores.  —  If,    in    formula 
.),  for    /,     F    and     r,     there  be  substituted  their  proper 


values,  namely,    /=  320  +  —    (form.  233  •),    F=  62000    and 
r  —  0-04,     as  in  Arts.  367  and  368,  we  shall  have 

320  + 


(249.) 


95232 

which  is  a  rule  for  rolled-iron  headers   in   the  floors  of  first- 
class  stores. 


CARRIAGE   BEAM    WITH   ONE   HEADER.  351 

As  this  expression  is  the  same  as  (248.),  excepting  the 
numerical  coefficients,  the  example  of  the  last  article  will 
suffice  to  illustrate  it,  by  simply  substituting  the  coefficient 

y  y 

320  +  £       140  +  - 

- — —  in  place  of 

95232  71424 


5(3,  —  Carriage  Beam  with  One  Header.  —  Formula 
is  appropriate  for  a  case  of  this  kind,  but  it  is  for  a  beam  of 
rectangular  section.  To  modify  it  for  use  in  this  case,  we 
have  (205.)  I  =  ^bd*  ;  or  I2/  =  bd\  Substituting  for 
bd*,  in  (161.),  this  value,  we  have 

fmn  (ng+^cl')  —  \2lFr 


which  is  a  general  rule  for  this  case. 


514. — Carriage  Beam  with  One   Header,  for  Dwellings 

V 

etc. — In  formula  (250.),   putting   for    f    its  value     140  +  —- 

(form.  232^  for    F    its  value     62,000     (Table  XX.),  and  for 
r    its  value    0-03     (Art.  314),  we  have 


— -  1  mn 


/= 


12  x  62000  xo-03 


22320 

which  is  a  rule  for  the  moment  of  inertia  of  a  rolled-iron  car- 
riage beam,  with  one  header,  in  floors  of  assembly  rooms, 
banks,  etc.  With  the  moment  of  inertia  found  by  this  rule, 
the  required  beam  may  be  selected  from  Table  XVII. 


352  ROLLED-IRON   BEAMS.  CHAP.   XIX. 

515. — Example. — In  a  dwelling  floor  of  Paterson  9  inch 
70  pound  beams,  20  feet  long  and  2T8¥  feet  from  centres : 
Of  what  size  should  be  a  carriage  beam  which  at  5  feet 
from  one  end  carries  a  header  17  feet  long,  with  tail  beams 
1 5  feet  long  ? 

Here  7=70,  <r=2-8,  m  =  5,  n  =  15,  £=17  and 
/—  20;  and  by  (251.)  we  have 


140 

'22320'  05  x  17  +  f  x  2-8  x  20)  =  161  -990 

or  the  moment  of  inertia  required  is  161-990. 

By  reference  to  Table  XVII.  we  find  /=  159-597  as  the 
moment  of  inertia  of  the  9  inch  135  pound  Pittsburgh 
beam,  being  nearly  the  amount  called  for.  If  the  construc- 
tion of  the  floor  permit  the  use  of  a  beam  i-J  inches  higher, 
then  it  would  be  preferable  to  use  for  this  carriage  beam  one 
of  the  Trenton  zoj  inch  90  pound  beams;  as  these  beams, 
although  stronger  than  we  require,  are  yet  (being  45  pounds 
lighter)  more  economical. 

516. — Carriage  Beam  with  One  Header,  for  First -class 
Stores. — If,  in  formula  (250.),  f  be  substituted  by  its  value 

520+— (form.  233. \    F    by  its  value    62,000   (Table  XX.), 
and    r    by    0-04    (Arts.  367  and  368),  we  shall  have 


I  = 


(320  +  —\mn 
V         W 

1 2  X  62OOO  X  O  •  04 

320 +•—]*** 


which  is  a  rule  for  the  moment  of  inertia  for  rolled-iron  car- 
riage beams,  carrying  one  header,  in  first-class  stores. 


CARRIAGE   BEAM   WITH    TWO   HEADERS.  353 

517. — Example. — Of  what  size,  in  a  first-class  store, 
should  be  a  rolled-iron  carriage  beam  25  feet  long,  which 
carries  at  5  feet  from  one  end  a  header  20  feet  long,  with 
tail  beams  25  feet  in  length  ;  the  tail  beams  being  Trenton 
12^  inch  125  pound  beams,  placed  2f  feet  from  centres? 

Here  7—125,  c  —  2f,  m  —  5,  n  =  20,  g  =  20  and 
I  —  25  ;  and  by  formula  (252.)  we  have 


i  x  5  x  2O 


f==~  29?60  -  X  (20X20  +  1  X2|X  25)^545.090 


or  the  moment  of  inertia  required  is    545-090. 

To  supply  the  strength  needed  in  this  case,  we  may  take 
a  Trenton  io|  inch  135  pound  beam,  with  one  of  their 
I2j  inch  125  pound  beams;  as  these  two  bolted  together 
will  give  a  moment  of  inertia  a  little  less  than  the  com- 
puted amount.  It  will  be  more  economical,  however,  to  take 
two  of  the  12  inch  125  pound  beams,  since  the  weight  of 
metal  will  be  less,  although  the  strength  will  be  greater  than 
required. 


518.—  Carriage  Beam  with  Two  Headers  and  Two  Sets 
of  Tail  Beams. — Formula  (170  )  contains  the  elements  appro- 
priate to  this  case,  but  is  for  beams  of  rectangular  section. 
It  is  quite  general  in  its  application,  although  somewhat 
complicated.  A  more  simple  rule  is  found  in  formula  (174.). 
This  is  not  quite  so  general  in  application,  but  still  suffi- 
ciently so  to  use  in  ordinary  cases  (see  Art.  402),  In  any 
event,  the  result  derived  from  its  use,  if  not  accurate,  devi- 
ates so  slightly  from  accuracy  that  it  may  be  safely  taken. 
We  will  take,  then,  formula  (17 4-)  and  modify  it  as  required 


354  ROLLED-IRON   BEAMS.  CHAP.  XIX. 

for  the  present  purpose.     For     bd*     putting     I2/,     its  value 
(form.  205.),  we  have 

1 2lFr  —  fm  [%cnl+g  (mn  +  r )] 

)]        (253.) 


which  is  a  general  rule  for  the  case  above  stated  (see  Arts. 
I53  and  243). 


5 1 9. — Carriage  Beam  with  Two  Header§  and  Two  Set* 
of  Tail  Beam*,  for  Dwellings,   etc. — If,    in   formula    (253. \ 

140  +  —    be  substituted    for   /   (form.  232.),    62,000    for    F 

(Table  XX.),  and    0-03    for   r    (Art.  314),  then  we  have  as  a 
result 

140  +  -- 


which    is  a    rule  for  the  moment  of  inertia  for  this  case  as 
above  stated  (see  Arts.  153  and  243). 


520. — Example. — In  a  dwelling  having  a  floor  of  Pater- 
son  loj  inch  105  pound  rolled-iron  beams,  20  feet 
long,  and  placed  5  •  84  feet  from  centres :  Which  of  the  beams 
of  Table  XVII.  would  be  appropriate  for  a  carriage  beam  to 
carry  two  headers  16  feet  long,  one  located  9  feet,  and 
the  other  1 $  feet,  both  from  the  same  end  of  the  carriage 
beam?  (See  Arts.  153  and  243.) 

Here  the  two  headers  are  respectively  9  feet  and  5 
feet  from  the  walls.  The  one  9  feet  from  its  wall,  being 
farther  away  than  the  other,  will  create  the  greater  strain, 


CARRIAGE   BEAM   WITH   TWO   HEADERS,  FOR   STORES.      355 

and  therefore      m  =  9,       n  =  n,       r  —  15,      J  =  5,      /—  20, 
r  =  5-84     and    y  —  105  ;     and  by  formula  (254-}  we  have 


105 

te?+jjm 


or  the  required  moment  of  inertia  is  211-337.  By  reference 
to  Table  XVII.,  we  find,  as  the  nearest  above  this  amount,  the 
Trenton  or  Paterson  10^  inch  135  pound  beam,  of  which 
/  =  241-478,  and  which  will  be  the  proper  beam  for  this  case. 


521.— Carriage  Beam  with  Two  Header§  and  Two  Sets 
of  Tail  Beams,  for  First-elass  Stores. — If,  in  formula  (253.), 
there  be  substituted  for  F  its  value  62,000  (Table  XX.), 

v 
for   f    its  value     320  H —    (form.  233.),  and  for    r   its  value 

0-04    (Arts.  367  and  368),  we  shall  have 


y 

320  +  - 


7  = 


which   is  a  rule  for  the  moment  of  inertia  required  in  this 
case,  as  above  stated  (see  Arts.  153  and  '243). 


522. — Example. — In  a  store  having  a  floor  of  Trenton 
inch  150  pound  rolled-iron  beams,  25  feet  long  and  4-87 
feet  from  centres :  What  ought  a  carriage  beam  to  be  which 
carries  two  headers  20  feet  long,  one  located  10  feet  from 
one  wall,  and  the  other  at  7  feet  from  the  other  wall  ? 

Here  the  distances  to  the  header  more  remote  from  its 
wall   are   to   be  called   (see  Arts.  153   and  243)    m    and    n. 


356  ROLLED-IRON   BEAMS.  CHAP.    XIX. 

Then      m  =  10,      »  =  15,      r  =  18,      5  =  7,     ^=20,      /=25, 
r  =  4-87     and    7  —  150;     and  by  formula  (255.) 

1 50 
320  +  —-^- 


4-87  x  15  x  25)  + 20(10  x  15+7')] 
=  695027 

or  the  moment  required  is  695027.  By  an  examination  of 
Table  XVII. ,  we  find  that  the  moment  of  the  Trenton  15^ 
inch  200  pound  beam  is  more  than  enough  for  this  case, 
and  its  use  more  economical  than  any  combination  of  other 
beams  affording  the  requisite  strength. 

523. — Carriage  Beam  with  Two  Headers,  Equidistant 
from  Centre,  and  Two  Sets  of  Tail  Beams,  for  Dwellings, 
etc. — If  for  /,  F,  bd3  and  r,  in  formula  (183.),  their  re- 

I/ 

spective  values  be  substituted,  namely,    /=  140+  —  (form. 


>.),     F=  62000     (Table   XX.),     bd3  =  I2/     (form.  205.  \ 
and     r  —  0-03     (Art.  314);  then  formula  (183.)  becomes 


y 
140  + 

-  ¥ 
22320 


(256.) 


which  is  a  rule  for  a  rolled-iron  carnage  beam,  carrying  two 
headers  equidistant  from  the  centre,  with  two  sets  of  tail 
beams,  in  assembly  rooms,  banks,  etc. 

524.  —  Example.  —  In  an  assembly  room,  having  a  floor  of 
Buffalo  io£  inch  105  pound  rolled-iron  beams,  20  feet 
long  and  5-35  feet  from  centres:  What  ought  a  carriage 
beam  to  be  which  carries  two  headers  16  feet  long,  located 
equidistant  from  the  centre  of  the  width  of  the  floor,  with  an 
opening  between  them  6  feet  wide  ? 


CARRIAGE  BEAM  WITH  TWO   EQUIDISTANT  HEADERS.     357 

Here    ^=5.35,    jj/ =  105,     /  =  20,    g  =  16    and     m  =  7  ; 
therefore  by  formula  (256.)  we  have  » 

105 

140  ' 


2232Q5'35  x  20 (Ax  5-35x20'+ 16x7*)=  190761 

By  reference  to  Table  XVII.  we  find  that  either  the 
Paterson  or  Trenton  ioj  inch  105  pound  beam  is  sufficiently 
strong  to  serve  for  the  required  carriage  beam. 


525. — Carriage  Beam  with  Two   Headers,  Equidistant 
from  Centre,  and  Two  Sets  of  Tail  Beams,   for  First-elass 

Stores.  — In  formula  (183^  if  we  substitute  for    /,    F,    bd3 

v 
and     r     their   respective  values,  as   follows,    f=  320  +  —-- 

(form.   233.),     F  =  62000     (Table  XX.),     bd9  =  12!    (form. 
205.)  and     r  =  0-04    (Arts.  367  and  368),  we  shall  have 


320+ 


which  is  a  rule  for  a  rolled-iron  carriage  beam  carrying  two 
headers  equidistant  from  the  centre,  with  two  sets  of  tail 
beams,  in  first-class  stores. 


526. — Example. — In  a  first-class  store,  having  a  floor  of 
Phoenix  15  inch  150  pound  beams  25  feet  long  and  4-75 
feet  from  centres :  What  ought  a  carnage  beam  to  be  which 
carries  two  headers  20  feet  long,  located  equidistant  from 
the  centre  of  the  width  of  the  floor,  with  an  opening  between 
them  8  feet  wide  ? 


358  ROLLED-IRON   BEAMS.  CHAP.  XIX. 

Here  we  have    ^=150,     £  =  4-75,     /=2$,    g—  20     and 
m  =  8^;     therefore  formula  (257.)  becomes 


320 


3  x  475 


29760 

or  the  moment  required  is  658-813.  Table  XVII.  shows 
that  either  of  the  15  inch  200  pound  beams  is  of  sufficient 
strength  to  satisfy  the  requirements  of  this  case. 


527. — Carriage  Beam  with  Two  Headers  and  One  Set 
of  Tail  Beams,  for  Dwellings,  etc. — If,  in  formula  (179.),  we 
substitute  for  the  symbols  bd3,  /,  F  and  r,  their  respec- 

y 

tive  values,  as  follows,    bd3  =  12!    (form.  205.),    f  —  140  +  -; 

6C 

(form.  232.),  F  =  62000  (Table  XX.)  and  ^'  =  0-03  (Art. 
314),  we  shall  have 

140  +  — 
I=     22320*  m  H™/+£>  (»  +  *M         (258-) 

which  is  a  rule  for  the  moment  of  inertia  of  a  rolled-iron 
carriage  beam,  carrying  two  headers  with  one  set  of  tail 
beams,  for  floors  of  assembly  rooms,  banks,  etc.  (See  Arts. 
153  and  409.) 


528, — Example. — In  a  bank  having  a  floor  of  Paterson 
loj-  inch  105  pound  rolled-iron  beams,  20  feet  long  and 
5  •  84  feet  from  centres  :  What  ought  a  carriage  beam  to  be 
which  carries  two  headers  16  feet  long,  located  one  at  5 
feet  from  one  wall  and  the  other  at  6  feet  from  the  other 
wall,  the  tail  beams  being  between  them? 


CARRIAGE   BEAM   WITH    TWO    HEADERS,    FOR   STORES.     359 

Here  (Art.  157)  m  is  to  be  put  at  the  wider  opening, 
hence  m  —  6,  71=14,  s  =  $,  I  —  20,  ^=5-84,  £•=  16, 
y  =  /—(/#+<$)  =  20—  ii  ="9  and  jy  =  105  ;  and  by  formula 

(258.) 

105 


/=  --  22320  *  '84x6[fx5>84xi4x20+i6x9x(i4  +  5)] 

=  187-593 

or,  the  moment  required  is  187-593.  Referring  to  Table 
XVII.  we  find  that  either  the  Paterson  or  Trenton  loj  inch 
105  pound  beam  will  be  suitable  for  this  case. 


529, — Carriage  Beam  with  Two  Headers  and  One  Set 
of  Tail  Beams,  for  First-class  Stores. — If,  in   formula    (258), 

y  y 

320  +  --         140  +  -^- 

we  substitute  (as  in  Art.  525)     —   —^—    for ~      we 

29760  22320 

shall  have 

320  +  -^- 

7  =  '  m 


which  is  a  rule  for  the  moment  of  inertia  for  a  rolled-iron 
carriage  beam,  carrying  two  headers  and  one  set  of  tail 
beams,  in  a  first-class  store. 


530. — Example. — In  a  first-class  store  having  a  floor  of 
Buffalo  15  inch  150  pound  beams  25  feet  long  and  4^ 
feet  from  centres :  What  ought  a  carriage  beam  to  be  which 
carries  two  headers  20  feet  long,  located,  one  at  5  feet 
from  one  wall,  and  the  other  at  8  feet  from  the  other  wall, 
with  tail  beams  between  them  ? 


360  ROLLED-IRON   BEAMS.  CHAP.  XIX. 

Here  (Art.  157)     m  =  8,     n  =  17,     s  =  5,     /=  25,    £•=  20, 
/=  25 -(5  +8)=  12,    ^  =  4i     and    j>  =  150;  and  by  (269.) 


320  + 


/=  " 


29760 

=  682-750 

which  is  the  moment  required.     Either  of  the     15    inch    200 
pound  beams  of  Table  XVII.  will  serve  the  present  purpose. 


531. — Carriage  Beam  with  Three  Headers,  the  Oreate§t 
Strain  being  at  Outside  Header,  for  Dwellings,  etc. — As  in 

Fig.  54,  floor  beams  are  sometimes  framed  with  two  openings, 
one  for  a  stairway  at  the  wall,  and  another  for  light  at  or  near 
the  middle  of  the  floor.  In  this  arrangement  the  carriage 
beams  are  required  to  sustain  three  headers.  Formula  (190.) 
in  Art.  425  is  appropriate  to  this  case,  but  is  adapted  to  a 
beam  of  rectangular  section.  Substituting  for  bd3  its  value 

I2/    (form.  205),   for    /    its  value     140 +  •—    (form.  232), 

for  F  its  value  62,000  (Table  XX.),  and  for  r  its  value 
0*03  (Art.  314),  we  have 


140  +  — 
7  =  ~  22320*  m  \i™l+g(™n+s*-v>}-\      (260) 


which  is  a  rule  for  the  moment  of  inertia  for  a  rolled-iron 
carriage  beam  carrying  three  headers,  in  an  assembly  room, 
bank,  etc. ;  the  headers  placed,  as  in  Fig.  54,  so  that  the  one 
causing  the  greatest  strain  shall  not  be  between  the  other 
two.  (See  Arts.  252  to  254.) 


CARRIAGE   BEAM   WITH   THREE   HEADERS.  361 

532. — Example. — In  an  assembly  room  having  a  floor  of 
Trenton  9  inch  70  pound  beams  20  feet  long  and  2-80 
feet  from  centres :  Of  what  size  should  be  a  carriage  beam 
carrying,  as  in  Fig.  54,  three  headers  1 5  feet  long  ;  two  of 
them  located  at  the  sides  of  an  opening  6  feet  wide,  which 
is  placed  at  the  middle  of  the  width  of  the  floor,  and  the 
other  header  located  at  3  feet  from  one  of  the  side  walls  ? 

As  two  of  these  headers  are  equidistant  from  the  centre 
of  the  floor,  the  one  carrying  the  longer  tail  beams  will  pro- 
duce the  greater  strain  upon  the  carriage  beam  (Art.  253). 
The  distances  from  this  header,  therefore,  are  to  be  desig- 
nated by  m  and  ;/  (Art.  244-),  while  r  and  s  are 
to  represent  the  distances  from  the  other,  and  v  and  u 
are  to  be  the  distances  from  the  third  header ;  the  one  at  the 
stairway. 

Here  m  —  7,  w  =  13,  s  =  7,  t>=3,  /=2O,  ^=15, 
c  —  2  -  8  and  y  =  70  ;  and  by  formula  (260.)  we  have 


70 

140  +      ' 

7  = 22/20  2  ~  x  7  [(*  x  2'8  x  13  x  20)  +  15 

=  133746 


which  is  the  required  moment.  An  examination  of  Table 
XVII.  shows  that  the  Trenton  9  inch  125  pound  beam 
will  be  more  than  sufficient  for  this  case. 


533. — Carriage  Beam  with  Three  Headers,  the  Greatest 
Strain   being   at   Outside    Header,  for    First-class   Stores. — 

Here,  with  the  headers  located,  as  in  Fig.  54,  so  that  the  one 
causing  the  greatest  strain  in  the  carriage  beam  shall  not  be 
between  the  other  two,  the  rule  is  the  same,  with  the  excep- 


362  ROLLED-IRON   BEAMS.  CHAP.   XIX. 


tion  of  the  coefficient,  as  in  the  case  last  presented  (form. 
£60.).     Substituting  therefore,  in  formula  (260.), 


320+-- 


(see  form.  259.)  in  place  of    ---  ~L      we  shall  have 

22320 


320  +  -- 
1=  n  \%cnl+g(mn  +  J-v*)-\     (261.) 


which  is  a  rule  for  the  moment  of  inertia  required  for  a 
rolled-iron  carriage  beam  carrying  three  headers,  in  a  first- 
class  store  *  the  headers  being  placed,  as  in  Fig.  54,  so  that 
the  one  carrying  the  greatest  strain  shall  not  be  between  the 
other  two.  (See  Arts.  252  to  254.) 

The  example  given  in  Art.  532  will  serve  to  illustrate  this 
rule,  for  the  two  rules  are  alike  except  in  the  coefficient,  as 
above  explained. 

534. — Carriage  Ream  with  Three  Headers,  the  Greatest 
Strain  being  at  Middle  Header,  for  Dwellings,  etc. — If  the 

headers  be  located  as  in  Fig.  .56,  so  that  the  header  causing 
the  greatest  strain  in  the  carriage  beam  shall  be  between  the 
other  two  (Arts.  260  and  264),  then  we  have  formula  (194-) 
(in  Art.  432)  appropriate  to  this  case,  except  that  it  is  for  a 
beam  of  rectangular  section.  To  modify  it  to  suit  our  pres- 
ent purpose,  we  have  only  to  substitute  for  bd8,  /,  F  and 
r,  their  respective  values  as  in  Art.  531,  and  we  have 


1=  -^\?*(b*l+g?)+gn(m*-it)-]     (262.) 
as  a  rule  for  the  moment  of  inertia  required  for  rolled-iron 


CARRIAGE  BEAM  WITH  THREE  HEADERS.       363 

carriage  beams  carrying  three  headers,  in  an  assembly  room, 
etc. ;  the  headers  so  located  that  the  one  causing  the  greatest 
strain  shall  be  between  the  other  two.  (See  Art.  264.) 


535. — Example. — In  a  bank,  having  a  floor  of  Phcenix 
lo^  inch  105  pound  rolled-iron  beams,  20  feet  long  and 
placed  5  -  59  feet  from  centres :  Of  what  size  ought  a  car- 
riage beam  to  be  which  carries  three  headers,  16  feet  long, 
placed,  as  in  Fig.  54,  so  that  the  opening  in  the  floor  at  the 
wall  shall  be  4  feet  wide,  and  the  other  opening  5  feet 
wide,  and  distant  6  feet  from  the  other  wall  ? 

The  middle  header  in  this  case  being  the  one  which 
causes  the  greatest  strain  in  the  carriage  beam,  the  distances 
from  it  to  the  two  walls  are  to  be  called  m  and  n.  (See 
Arts.  244  and  253.)  The  header  carrying  the  tail  beams, 
one  end  of  which  rest  upon  the  wall  causing  the  next  great- 
est strain,  the  distances  from  it  to  the  walls  are  to  be  called 
r  and  s.  The  distances  from  the  third  header  are  v  and 
u.  We  have,  therefore,  m  =  9,  n  =  11,  s  =  6,  v  —  4, 
1—20,  g  —  16,  y  —  105  and  ^=5-59;  and  by  formula 
(262.}  have 


105 
140  +  - 


5-59 


16(11  x  c/-4*)]=  199-597 

or  the  required  moment  is  199-597.  From  the  moments  in 
Table  XVII.  we  find  that  the  Phcenix  and  Pittsburgh  loj 
inch  135  pound  beams  are  a  trifle  stronger  than  the  required 
amount.  The  Trenton  and  Paterson  ioj  inch  135  pound 
beams  are  still  stronger  than  this.  Being  of  the  same  weight, 
either  of  the  four  named  beams  will  serve  the  purpose. 


364  ROLLED-IRON   BEAMS.  CHAP.    XTX. 

536.—  Carriage  Beam  with  Three  Headers,  the  Greatest 
Strain  being  at  middle  Header,,  for  First-clas§  Stores.  —  Take 
a  case  where  the  header  causing  the  greatest  strain  in  the 
carriage  beam  occurs  between  the  other  two,  as  in  Fig.  56. 
Formula  (262.)  is  suitable  for  this  case,  except  in  its  coeffi- 

320  +  J 
cient.     To  modify  it   to  suit   our  purpose,  let    -  g~-    in 

140  +  — 

formula  (261.)  be  substituted  for  -  -  —   in  formula  (262.)] 

22320 

and  we  have 

y 
320  + 


-^ 

which  is  a  rule  for  the  moment  of  inertia  for  rolled-iron  car- 
riage beams  carrying  three  headers,  in  first-class  stores  ;  the 
header  causing  the  greatest  strain  being  between  the  other 
two.  (See  Art.  264.) 

The  example  given  in  Art.  535  will  be  sufficient  to  illus- 
trate this  rule,  as  the  two  formulas  are  alike,  except  in  their 
coefficients. 


QUESTIONS   FOR   PRACTICE. 


537. — What  is  the  moment  of  inertia  for  a  beam  having 
a  rectangular  section  ? 

538. — What  is  the  moment  of  inertia  for  a  beam  of 
I  section,  or  of  the  form  of  rolled-iron  beams  ? 

539._Which  of  the  beams  of  Table  XVII.  would  be  ap- 
propriate, when  laid  upon  two  supports  25  feet  apart,  to 
sustain  15,000  pounds  at  the  middle,  with  a  deflection  of 
f  of  an  inch  ? 

54-0. — What  weight  could  be  sustained  at  10  feet  from 
one  end  of  a  Trenton  10^  inch  105  pound  beam,  25  feet 
long  between  bearings,  with  a  deflection  of  one  inch  ? 

541. — What  weight  uniformly  distributed  could  be  sus- 
tained upon  a  Buffalo  9  inch  90  pound  beam,  projecting* 
as  a  lever  1 5  feet  from  a  wall  (in  which  one  end  is  firmly 
imbedded),  with  a  deflection  of  \  an  inch  ? 

542. — In  the  floors  of  a  first-class  store,  constructed  with 
Phoenix  12  inch  125  pound  beams,  3^  feet  from  centres : 
Which  of  the  beams  of  Table  XVII.  ought  to  be  used  for  a 
header  15  feet  long,  carrying  one  end  of  a  set  of  tail  beams 
12  feet  long  ? 


366  ROLLED-IRON    BEAMS.  CHAP.    XIX. 

543. — In  the  floor  of  a  first-class  store,  constructed  with 
12  inch  125  pound  beams  2\  feet  from  centres :  Which  of 
the  beams  of  Table  XVII.  ought  to  be  used  for  a  carriage 
beam  25  feet  long  between  bearings,  carrying,  with  0-04 
of  an  inch  per  foot  deflection,  a  header  20  feet  long,  located 
at  7  feet  from  one  end  of  the  carriage  beam,  and  carrying 
one  end  of  a  set  of  tail  beams  18  feet  long? 

54-4. — In  the  floor  of  a  first-class  store,  constructed  of 
15  inch  150  pound  beams  4^  feet  from  centres :  What  size 
should  be  a  carriage  beam  25  feet  long,  which  carries  two 
headers  19  feet  long,  one  located  at  9  feet  from  one  wall, 
and  the  other  at  8  feet  from  the  other  wall ;  the  two  head- 
ers having  an  opening  between  them  ? 

545. — In  the  floor  of  a  bank,  constructed  of  loj  inch 
105  pound  beams  22  feet  long,  and  placed  4  feet  4 
inches  from  centres :  Of  what  size  should  be  a  carriage  beam 
which  carries  three  headers,  16  feet  long,  and  located,  as  in 
Fig.  56,  so  that  one  opening  at  the  wall  shall  be  3  feet  wide, 
and  the  other  opening  6  feet  wide,  with  a  width  of  floor  of 
6  feet  between  the  two  openings  ? 


CHAPTER    XX. 


TUBULAR  IRON   GIRDERS. 

ART.  546. — Introduction  of  the  Tubular  Girder. — Dur- 
ing the  construction  of  the  great  tubular  bridges  over  the 
Con  way  River  and  the   Menai  Straits,  Wales  (1846  to  1850), 
engineers  and   architects  were   moved  with  new  interest  in 
discussions  and  investigations  as  to  the  possibilities  of  con- 
structions involving  transverse  strains.     Since  the  complete 
success  of  those  justly  celebrated  feats  of  engineering  skill, 
the  tubular  girder  (Fig.  74),  as  also  the  plate  girder  (Fig.  67), 
and  the  rolled-iron  beam  (Fig.  68), 
all  of  which  owe  their  utility  to 
the  same  principle  as  that  involv- 
ed in  the  construction  of  the  tu- 
bular girder,  have  become  deserv- 
edly   popular.       They    are    now 
extensively  used,  not  only  by  the 
engineer   in    spanning    rivers  for 
the  passage  of  railway  trains,  but 
also  by  the  architect  in  the  lesser, 
but    by    no    means    unimportant, 
work  of  constructing  floors  over  FlG-  74- 

halls  of  the  largest  dimensions,  without  the  use  of  columns 
as  intermediate  supports. 


547.— Load  at   Middle— Rule    Essentially   the   Same    as 
that  for  Rolled-Iron  Beams. — The  capacity  of  tubular  gir- 


TUBULAR   IRON    GIRDERS.  CHAP.    XX. 

ders  may  be  computed  by  the  rules  already  given.  For 
example  :  Formula  (216.)  affords  a  rule  for  a  load  at  the 
middle  of  a  rolled-iron  beam,  in  which  (form.  213.), 


whereof  b  is  the  width  of  top  or  bottom  flange,  and  bt 
equals  b,  less  the  thickness  of  the  two  upright  parts,  or  webs  ; 
d  is  the  entire  depth,  and  d,  is  the  depth,  or  height,  in  the 
clear  between  the  top  'and  bottom  flanges,  bd  then  is  the 
area  of  the  whole  cross-section,  measured  over  all,  while  btdt 
represents  the  area  of  the  vacuity,  or  of  so  much  of  the  cross- 
section  as  is  wanting  to  make  it  a  solid.  The  numerical  coef- 
ficient in  formula  (216)  is  based  upon  a  value  of  F  equal  to 
62,000,  which  is  the  amount  derived  from  experiments  on 
solid  rolled-iron  beams.  For  built  beams,  such  as  the  tubular 
girder,  F  by  experiment  would  prove  to  be  less,  but  the 
formula  (216)  may  be  used  as  given,  provided  that  proper 
allowance  be  made  in  the  flanges  on  account  of  the  rivet 
holes  ;  that  is,  taking  instead  of  the  actual  breadths  of  the 
flanges  only  so  much  of  them  as  remains  uncut  for  rivets. 


548. — Load  at  Any  Point— Load  Uniformly  Distributed. 

—For  a  load  at  any  point  in  the  length  of  a  beam,  formula 
(222.)  will  serve,  while  for  a  load  uniformly  distributed,  for- 
mula (228)  affords  a  rule.  In  general,  any  rule  adapted  to 
rolled-iron  beams  will  serve  for  the  tubular  or  plate  girder, 
by  taking  as  the  areas  of  metal  the  uncut  portion  only. 


54-9. — Load  at  Middle— Common  Rule. — The  rules  just 
quoted  are  not  those  which  are  generally  used  for  tubular 
beams.  Preliminary  to  planning  the  Conway  and  Britannia 


LOAD   AT   MIDDLE.  369 

tubular  bridges,  the  engineers  tested  several  model  tubes, 
and  from  them  deduced  the  formula 


in  which    C  is  a  constant,  found  to  be  equal  to   80   when   W 
represents  gross  tons.     Changing    W   to  pounds,  we  have 

2240  x  80  x  a'd  a'd 

W—-       —j—       -=i792oo-y- 

This  is  for  the  breaking  weight.  Taking  the  safe  weight 
as  9000  pounds  per  inch,  or  £  of  the  breaking  weight,  we 
have 

-  =  35840 

and,  as  an  expression  for  the  safe   weight,  the  area  of  the 
bottom  flange  equals 

Wl 


a   = 


35840^ 


or,  if  instead  of  the  above  constant,    80,   we  put   80-357,   we 
shall  have  our  constant  in  round  numbers,  thus, 


Wl 

a'  = 


which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  tubular 
girder,  with  the  load  at  the  middle;  a'  being  in  inches,  / 
and  d  in  feet,  and  W  in  pounds.  This  rule  is  identical 
with  formula  (265.),  deduced  in  another  manner. 

550. — Capacity  by  the  Principle  of  Moments. — Gene- 
rally, the  strength  of  tubular  beams  is  ascertained  by  the 
principle  of  moments  or  leverage.  Sufficient  material  must  be 


37°  TUBULAR   IRON    GIRDERS.  CHAP.    XX. 

provided  in  the  top  flange  to  resist  crushing,  and  in  the  bot- 
tom flange  to  resist  tearing  asunder,  while  the  material  in  the 
web  or  upright  part  should  be  adequate  to  resist  shearing. 


551.  —  Load  at  Middle—  Momeiat§.  —  We  will  first  con- 
sider the  requirements  in  the  flanges. 

The  leverage,  or  action  of  the  power  tending  to  break  the 
beam,  as  also  that  of  the  resistance  of  the  materials,  is  repre- 
sented in  Figs.  8  and  9.  When  the  load  upon  a  beam  is  con- 
centrated at  the  middle,  it  acts  with  a  power  of  half  the 
weight  into  half  the  length  of  the  beam  (Art.  35),  and  the 
tension  thereby  produced  in  the  bottom  flange  is  resisted 
by  a  leverage  equal  to  the  height  of  the  beam  ;  or,  if  d 
equals  the  height  of  the  beam  between  the  centres  of  gravity 
of  the  cross-sections  of  the  top  and  bottom  flanges,  and  T 
equals  the  amount  of  tension  produced  in  the  lower  flange 
by  the  action  of  a  weight  W  upon  the  middle  of  the  beam, 
then 


Again,  if  k  equals  the  pounds  per  square  inch  of  section 
with  which  the  metal  in  the  lower  flange  may  be  safely 
trusted,  and  a'  equals  the  area  in  inches  in  the  bottom 
flange,  then  a'k  =  T,  and 

=  a'kd 
Wl 


which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  tubular 
girder,  loaded  at  the  middle,  and  in  which  W  and  k  are 
in  pounds,  a  is  in  inches,  and  d  and  /  are  in  feet.  (The 


COMPUTATION   BY    MOMENTS.  371 

area  of  the  top  flange  is  to  be  made  equal  to  that  of  the 
bottom  flange.  See  Art.  456.)  If  k  be  taken  at  9000, 
as  in  Art.  549,  then  4/£  =  36000,  and  formula  (265)  becomes 
identical  with  formula  (264). 

552. — Example. — What  area  of  metal  would  be  required 
in  the  bottom  flange  of  a  tubular  girder  40  feet  long  and 
3  feet  high,  to  sustain  at  the  middle  75,000  pounds;  9000 
pounds  being  the  weight  allowed  upon  one  inch  of  the 
wrought-iron  of  which  the  flanges  are  to  be  made  ? 

Here  W  =.  75000,  /  =  40,  d  —  3  and  k  —  9000  ;  and 
we  have,  by  formula  (265), 

75000  x  40 

"-27.77 


4  x  3  x  9000 

or  the  area  equals   27^   inches.     This  is  the  amount  of  metal 
in  addition  to  that  required  for  rivet  holes. 

553.  —  Load  at  Any  Point.  —  A  load  concentrated  at  any 
point  in  the  length  of  the  beam  acts  with  a  leverage  equal  to 

W  y-     (see   Art.    56),   and   the    resistance    is     Td=a'kd', 
therefore 


which  is  a  rule  for  this  case,  as  above  stated,  in  which    af   is 
in  inches,    W   is  in  pounds,  and   m,   n,   d  and   /  are  in  feet. 

554.  —  Example.—  What  amount  of  metal  would  be  re- 
quired in  the  bottom  flange  of  a  tubular  girder   50   feet  long 


372  TUBULAR  IRON   GIRDERS.  CHAP.   XX. 

and  3^  feet  high,  to  sustain  a  load  of  50,000  pounds  at  20 
feet  from  one  end,  when  k  =  9000  ? 

Here     W  —  50000,     ;;/  =  20,    n  =  30,     d  =  3^,    k  —  9000 
and     /  =  50  ;     and,  by  formula  (266.), 

50000  x  20  x  30 
*'=—•  —*-  =  19-05 

3^  x  9000  x  50 

or  the  area  should  have  19  inches  of  solid  metal,  uncut  by 
rivet  holes.  The  top  flange  should  contain  an  equal  amount. 
(See  Art.  456.) 


555.  —  Load  Uniformly  Distributed.  —  For  this  load  the 
effect  at  any  point  in  the  beam  is  equal  to  that  of  half  the 
load,  if  concentrated  at  that  point  (see  Art.  214);  or,  from 
formula 


which  is  a  rule  for  the  area  of  the  bottom  flange  at  any  point 
in  its  length,  and  in  which  a'  is  in  inches,  U  is  in 
pounds,  and  m,  n,  d  and  /  are  in  feet. 


556.  —  Example.  —  In  a  tubular  girder  50  feet  long,  3^ 
feet  high,  and  loaded  with  an  equably  distributed  load  of 
120,000  pounds  :  What  should  be  the  area  of  the  bottom 
flange  at  the  middle,  and  at  each  5  feet  of  the  length 
thence  to  each  support,  k  being  taken  at  9000  ? 

Here  U  =  1  20000,  d  —  3  £,  k  =  9000  and  /  =  50  ; 
and  by  formula  (267.)  we  have 


12OOOOMH 

a>  —. 


2  x  3J  *  9000  x  50 


UNIFORMLY   DISTRIBUTED   LOAD.  3/3 

When     m  =  n  —  25,     then 

a'  =  0-038095  x  25  x  25  =  23-81 

or  the  area  required  in  the   bottom  flange  at  mid-length  is 
23-81     inches. 

When     m  =  20,     then     n  =  30,     and 

a'  =  o •  038095  x  20  x  30  =  22  •  86 

or  the  required  area  at     5     feet  from  the  middle,  either  way, 
equals    22-J    inches. 

When     m=i$,     then     n  —  35,     and 

a'  =  0-038095  x  15  x  35  =  20-00 

or,  at     10    feet  each  side  of  the  middle,  the  area  should  be 
20    inches. 

When     m  =  10,     then     n  —  40,     and 

a'  =  0-038095  x  10  x  40  =  15-24 

or,  at     15     feet  each  side  of  the  middle,  the  area  should  be 
I5J-    inches. 

When     m  =  5,     then     n  =  45,     and 

a' =  0-038095  x5  x45  =  8-57 

or,  at     20    feet  each  side  of  the  middle,  the  area  should  be 
84    inches. 


557. — Thickness  of  Flanges. — In  the  results  of  the  ex- 
ample just  given,  it  will  be  observed  that  the  area  of  metal 
required  in  the  flanges  increases  gradually  from  the  points  of 
support  each  way  to  the  middle  of  the  beam  (see  Art.  178). 
In  practice,  this  requirement  is  met  by  building  up  the 
flanges  with  laminas  or  plates  of  metal,  lapping  on  according 


3.74  TUBULAR   IRON   GIRDERS.  CHAP.    XX. 

to  the  computed  necessary  amount.  In  this  process,  the 
plates  used  are  generally  not  less  than  J  of  an  inch  thick. 
For  an  example,  take  the  results  just  found.  Adding,  say  | 
for  rivet  holes,  and  dividing  the  sum  by  the  width  of  the 
girder,  which  we  will  call  12  inches,  there  results  as  the 
thickness  of  rnetal  required, 

at  the  middle,  2-31,  say  2^  inches ; 

"     5    feet  from  middle,  2-22,       "  2j      " 

"    10          "               "  1-95,       "  2 

"    15          "               "  1-48,       "  i£      " 

"   20          "                "  0-83,  "  i  inch. 


558. — €on§truetion  of  Flanges. — The  girder  of  the  last 
article  might  be  built  with  the  two  flanges  in  plates  12 
inches  wide,  thus :  Lay  down  first  a  plate  one  inch  thick 
the  whole  length  of  the  girder.  (With  an  addition  for  supports 
on  the  walls,  say  -fa  of  the  length,  or  2^  feet  at  each  end, 
this  plate  would  be  55  feet  long.)  Upon  this  place  a  plate 
£  inch  thick  and  40  feet  long  ;  on  this  a  plate  \  inch 
thick  and  30  feet  long;  on  this  a  plate  J  inch  thick 
and  20  feet  long;  and  on  this  a  plate  \  inch  thick  and 
10  feet  long.  The  plates  are  all  to  extend  to  equal 
length  each  side  of  the  middle  of  the  girder,  and  to  be  well 
secured  together  by  riveting.  The  longer  plates,  probably, 
will  have  to  be  in  more  than  one  piece  in  length.  Where 
heading  joints  occur,  a  covering  plate  should  be  provided 
for  the  joint  and  riveted. 


559. — Shearing  Strain. — A  sufficient  area  having  been 
provided  in  the  top  and  bottom  flanges  to  resist  the  com- 
pressive  and  tensile  strains,  there  will  be  needed  in  the  web 
metal  sufficient  to  resist  only  the  shearing  strain.  This  strain 


SHEARING   STRAIN.  375 

is,  theoretically,  nothing  at  the  middle  of  a  beam  uniformly 
loaded,  but  from  thence  increases  by  equal  increments  to 
each  support,  at  which  place  it  is  equal  to  one  half  of  the 
whole  load  (Arts.  !72  and  174).  For  example  :  In  the  case 
considered  in  Art.  556,  the  beam,  50  feet,  long,  carries 
120,000  pounds  uniformly  distributed  over  its  whole  length ; 
half  of  the  load  over  half  of  the  beam.  At  the  centre,  the 
shearing  strain  is  nothing ;  at  5  feet  from  the  centre,  it  is 
equal  to  -§-  of  half  the  load,  or  is  equal  to  12,000  pounds  ; 
at  10  feet  it  is  24,000;  at  15  feet  it  is  36,000;  at  20 
feet  it  is  48,000;  and  at  25  feet,  or  at  the  supports,  it  is 
60,000  pounds,  or  half  the  whole  load. 


560. — Thickiies§  of  Wefo. — If     G     be   put  for  the  shear- 
ing stress,  then 

G  =  a'k' 

in  which  a  is  the  area  in  inches  of  the  web  at  the  point 
of  the  stress,  and  k'  is  the  effective  resistance  of  wrought- 
iron  to  shearing,  per  inch  area  of  cross-section.  If  t  equals 
the  thickness,  and  d  the  height  of  the  web,  then  a'  =  td, 
and  the  above  equation  becomes 

G  =  k'td 

*  =  w 

which  is  a  rule  for  the  thickness  of  the  web,  at  any  point  in 
the  length  of  the  beam,  and  in  which  t  and  d  are  in 
inches. 


561.— Example.— What  should    be   the  thickness  of  the 
web  of  the  tubular  girder  considered  in  Art.  556,  computed 


376  TUBULAR   IRON   GIRDERS.  CHAP.    XX. 

at  every  5  feet  in  length  of  the  girder  ?  If  k'  be  taken  at 
7000  pounds,  it  will  be  but  little  more  than  three  quarters  of 
9000,  the  amount  taken  in  tensile  strain  (Art.  173),*  and 
taking  d  at,  say  38  inches,  we  have,  by  formula  (268.\ 


38  x  7ooo       266000 
Therefore,  when     G    equals    60,000    (Art.  559),   then 

60000 

t  =  -£z =0-225 

266000 

When    G   equals  48,000,    then   t  =    ^,     -  =  o- 180.     When 


G    equals     36,000,     then     £= -ig =  0-135.     As  these  are 

the  greater  of  the  strains,  and  are  all  below  the  practical 
thickness  in  girders,  it  is  not  worth  while  to  compute  those 
at  the  remainder  of  the  stations. 


562. — Conitruction  of  Web. — From  the  results  in  the 
last  article,  it  appears  that  in  this  case  the  web  is  required,  of 
necessity,  to  be  only  a  quarter  of  an  inch  thick  in  its  thickest 
part,  at  the  supports.  With  an  increase  of  load,  the  thick- 
ness of  the  web  would  increase,  for  by  the  formula  it  is 
directly  as  the  load. 

The  thickness  of  web  just  computed  is  the  whole  amount 
required  in  the  two  sides  of  the  girder.  In  practice,  it  is 
found  unwise  to  use  plates  less  than  a  quarter  of  an  inch 
thick.  Following  this  custom,  the  two  sides  of  the  girder 


*  The  resistance  to  shearing  is  generally  taken  at  three  quarters  of  the  ten- 
sile strength  (see  Haswell's  Engineers'  and  Mechanics'  Pocket  Book,  p.  485 — 
Weisbach's  Mechanics  and  Engineering,  vol.  2,  p.  77). 


CONSTRUCTION   OF  WEB.  377 

taken  together  would  be  half  an  inch  thick,  more  than  twice 
the  amount  of  metal  actually  required.  Hence  it  may  justly 
be  inferred  that  in  similar  cases  the  plate  beam  (Fig.  67) 
would  be  preferable  to  the  tubular  girder,  as  its  web,  being 
single,  would  require  only  half  the  metal  that  would  be  re- 
quired in  the  two  sides  of  the  tubular  girder.  It  is  also  pre- 
ferable for  the  reason  that  it  is  more  easily  painted,  and  thus 
kept  from  corrosion.  On  the  other  hand,  a  tubular  beam  is 
stiffer  laterally.  In  the  construction  of  the  web,  as  a  precau- 
tion to  prevent  buckling,  or  contortion,  it  is  requisite  to  pro- 
vide uprights  of  T  iron,  at  intervals  of,  say  3  feet  on  each 
side,  to  which  the  web  is  to  be  riveted. 


563.  —  Floor  Girder—  Area  of  Flange.  —  If  for  U  in  for- 
mula (267.),  there  be  substituted  its  value  in  a  floor,  c'fl, 
of  which  c'  is  the  distance  from  centres  between  girders, 
or  the  width  of  floor  sustained  by  the  girder,  /  is  the  length 
of  the  girder  between  supports  (both  in  feet),  and  /  is  the 
load  per  foot  superficial  upon  the  floor,  including  the  weight 
of  the  materials  of  construction,  then 


which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  tubular 
girder,  sustaining  a  floor,  and  in  which  a'  is  in  inches  and 
c't  m,  n  and  d  are  in  feet. 


664. — Weight  of  the  Girder. — In  estimating  the  load  to 
be  carried  by  a  girder,  the  estimate  must  include  the  weight 
of  the  girder  itself,  It  is  desirable  therefore  to  be  able  to 


378  TUBULAR   IRON   GIRDERS.  CHAP.    XX. 

measure  its  weight  approximately  before  its  dimensions 
have  been  definitely  fixed.  The  weight  of  a  tubular  girder 
will  be  in  proportion  to  its  area  of  cross-section  (which  will 
be  approximately  as  the  load  it  has  to  carry),  and  to  its 
length  (form.  265^)  ;  'or,  when  U  is  the  gross  load  to  be  car- 
ried, and  /  the  length  between  bearings,  then  the  weight 
of  the  girder  between  the  bearings  is 


in  which  n  is  a  constant,  and  U  is  the  whole  load,  includ- 
ing that  of  all  the  materials  of  construction.  The  value  of 
n,  when  derived  from  so  large  a  structure  as  that  of  the  tu- 
bular bridge  over  Menai  Straits,  is  about  600,  but  from  sev- 
eral examples  of  girders  from  35  to  50  feet  long,  in  floors 
of  buildings,  its  value  is  found  to  be  about  700.  For  our 
purpose,  then,  we  have  n  —  700.  If  for  U  we  put  its 
equivalent  c'fl,  as  in  Art.  563,  then 

(270.} 


7oo 

This  is  the  weight  of  so  much  of  the  girder  as  occurs  within 
the  clear  span  between  the  supports. 

565. — Weight  of  Girder  per  Foot  Superficial  of  Floor. — 

The  area  of  the  floor  supported  by  a  girder  is  c'l.  Dividing 
K  by  this,  the  quotient  will  be  /',  the  weight  of  the  girder 
per  foot  superficial  of  the  floor,  thus : 


j^X.     -72° -   fL 
J     ~  c'l  ~     c'l  ~~  700 

Now   /,    the  total  load  per  foot  superficial  of  the  floor,  com- 


WEIGHT   OF   GIRDER.  379 

prises  the  superimposed  load,  the  weight  of  the  brick  arches, 
etc.,  and  the  weight  of  the  girder  /';  and,  putting  m  for 
the  weight  of  all  else  save  that  of  the  girder,  we  have 


f  —  m  +/'  and,  from  the  above, 

1L~  (? 

700 

Im+fl 


=          = 
*         700  700 


*  700 

;oo/'  a=  lm 
700/-/7  =  lm 


f  = 


700—  / 

which  is  a  rule  for  ascertaining  the  weight  per  foot  superfi- 
cial of  the  floor  due  to  the  tubular  girder. 


566.  —  Example.  —  A  floor,  the  weight  of  which,  including 
that  of  the  superimposed  load,  is  140  pounds  per  superficial 
foot,  is  carried  upon  a  girder  50  feet  in  length  between  its 
bearings.  What  additional  amount  per  foot  superficial 
should  be  added  for  the  weight  of  the  girder? 

Here    /  —  50    and    m  —  140,    and  by  (27  1.\ 


_ 

700  —  50 

or  the  weight  to  be  added  for  the  girder  is     lof    pounds. 
Then      f=  m+f  —  140+  lof  =  150}      pounds. 


567. — Total  Weight  of  Floor  per  Foot  Superficial,  in- 
eluding  Girder. — In  the  last  article  m  represents  the  weight 
of  one  foot  superficial  of  a  floor,  including  the  load  to  be  car- 


380  TUBULAR   IRON   GIRDERS.  CHAP.   XX. 

ried ;  also,  f  represents  the  weight  due  to  the  girder ;  or, 
for  the  total  load,  f=m+f.  Using  for  /'  its  value  as 
in  formula  (27 1.)  we  have 

/—  m+f  = 


7oo— / 
/=  m(i  -f 7  ) 

J  \          700— // 


f=m 


700- 
700 


700— / 


and  for    m,    taking  its  value  as  given  in  formula  (232.\  it 
being  there  represented  by   f, 


which  is  the  value  of  f,  the  total  load  per  superficial  foot 
of  the  floors  of  assembly  rooms,  banks,  etc.,  to  be  used  in  the 
calculation  of  tubular  girders ;  and  taking  the  value  of  m, 
as  expressed  in  formula  (233.)  we  have 


/=( 


which  is  the  corresponding  value  of  f  for  the  floors  of  first- 
class  stores. 

568. — Girders  for  Floors  of  Dwellings,  etc. — If  in  formula 
(269.),  we  substitute  for  /  its  value  as  in  formula  (272.),  we 
shall  have 

/  y\    700        c'mn 

a'  =  (ijp  +  f-Jife-r?  x  — 


3<V700- 

which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  tubular 
girder,  supporting  the  floor  of  an  assembly  room  or  bank, 
and  in  which  a'  is  in  inches,  and  c,  /,  c',  m,  n  and  d 
are  in  feet. 


TO   SUPPORT  FLOORS   OF  DWELLINGS.  381 

569.  —  Example.  —  In  a  floor  ot  9  inch  70  pound 
beams,  4  feet  from  centres:  What  ought  to  be  the  area  of 
the  bottom  flange  of  a  tubular  girder  40  feet  long  between 
bearings,  2|  feet  deep,  and  placed  17  feet  from  the  walls, 
or  from  other  girders  ;  the  area  of  the  flange  to  be  ascer- 
tained at  every  five  feet  in  length  of  the  girder  ? 

Here   y  =  70,    c  =  4,    /=  40,    c'  =  17    and    d—  2f. 

Putting    k    at    9000    we  have,  by  (2  74-), 

,       (  70  \     700  17 

a  =  (  140  +  —  —  x  -  —  x  mn 

3x4/700-40       2X2f 


The  values  of   m    and    n    are  as  follows: 

At  the  middle,  m  =  20  and  n  =  20 

"       5    feet  from  middle,  m  —  15      "  n  =  25 

"      10      "        "  "  m  =  10      "  n  =  30 

"      1-5       "        "  "  m  =    5       "  n  =  35 

from  which  the  values  of    a'    are  as  follows: 

At  the  middle,  a'  =  0-05478  x  20  x  20  =  21  -91 

"      5    feet  from  middle,  a'  =  0-05478  x  15  x  25  =  20-54 

"     10      "        "  "  a'  —  0-05478  x  10  x  30  =  16-43 

"     15       "        "  "  a'  =  0-05478  x    5X35=    9-59 

These  are  the  areas  of  cross-section  of  the  lower  flange,  at  the 
respective  points  named.  The  top  flange  is  to  be  of  the  same 
size.  (See  Art.  456.) 

570.  —  Girder§  for  Floors  of  Fir§t-cla§§  Storci.  —  If,  in  for- 
mula (2  74,),  320  be  substituted  for  140  (see  form.  233.), 
we  shall  have 


7oo—( 


TUBULAR   IRON   GIRDERS.  CHAP.    XX. 

which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  tubular 
girder  in  a  first-class  store.  [The  area  of  the  upper  flange 
should  be  made  equal  to  that  of  the  bottom  flange  (Art. 
456).] 

As  this  rule  is  similar  to  (274-},  the  example  given  to 
illustrate  that  rule  will  suffice  also  for  this. 


571. — Ratio    of  Depth    to    Length,  in  Iron  CJirder§.— In 

order  that  the  requisite  strength  in  tubular  girders  may  be 
attained  with  a  minimum  of  metal,  the  depth  of  a  girder 
should  bear  a  certain  relation  to  the  length.  To  deduce  a 
rule  for  this  ratio  from  mathematical  considerations  purely, 
is  not  an  easy  problem.  Baker  in  his  work  on  the  Strength 
of  Beams,  p.  288,  discusses  the  subject  at  some  length.  No 
more  will  be  attempted  here  than  to  obtain  a  rule  based 
upon  some  general  considerations,  and  upon  results  tested 
and  corrected  by  experience. 

572. — Economical  Depth. — In  the  construction  of  tubular 
girders  for  the  floors  of  large  buildings,  it  is  found  in  practice* 
to  be  unadvisable  to  use  plates  of  a  less  thickness  than  one 
quarter  of  an  inch.  If  each  side  of  the  girder  be  a  quarter 
of  an  inch  thick,  then  the  least  thickness  for  the  web  (using 
this  term  technically)  is  a  half  inch.  This  is  more  than  is 
usually  found  necessary,  in  this  class  of  girders,  to  resist 
shearing  (Art.  562).  As  the  thickness  is  thus  fixed,  there- 
fore the  area  of  the  web  will  be  in  proportion  to  its  height, 
and  consequently  it  is  advisable,  in  so  far  as  the  web  is  con- 
cerned, to  have  the  depth  of  the  girder  small ;  but,  on  the 
other  hand,  as  the  area  of  the  flanges  is  inversely  propor- 
tional to  the 'depth  (see  form.  £65.),  a  reduction  of  the  flanges 
will  require  that  the  depth  be  increased.  The  cost  of  the 
girder  is  in  proportion  to  its  weight,  which  is  in  proportion 


ECONOMICAL  DEPTH.  383 

to  its  area  of  cross-section,  and  hence  the  desirability  of 
making  both  as  small  as  possible. 

The  area  of  the  flanges  is,  by  formula  (265.),  in  propor- 

Wl 
tion  to    -T7,    and,  as  before  shown,  the  area  of  the  web  will 

be    in    proportion   to   its    height ;    or   the    whole   area    will 

Wl 

be  in  proportion  to    -rr  +  d ;    and  the  problem  is  to  find  such 
dk 

a  value  of  d  as  will  make  this  expression  a  minimum. 
Putting  the  differential  of  this  equation  equal  to  zero,  we 
find  that  the  area  of  the  cross-section  of  the  beam  will  be  a 
minimum  when 

Wl 


in  which  x  is  a  constant,  to  be  derived  from  experience, 
and  which,  by  an  application  of  the  formula  to  girders  of 
this  class,  is,  when  the  weight  is  equally  distributed,  found 
to  be  equal  to  30.  This  reduces  the  formula  to 


and  when  for    U  its  value   c'fl    is  substituted 


which  is  a  rule  for  ascertaining  the  economical  depth  of  a 
tubular  girder  ;  a  rule  useful  in  cases  where  the  depth  is  not 
fixed  by  other  considerations. 


573. — Example. — In  a  floor  where  the  girders  are  50 
feet  long  and  placed  15  feet  from  centres,  and  where  the 
total  load  per  foot  superficial  is  155  pounds:  What  would 
be  the  most  economical  depth  for  the  girders  ? 


TUBULAR   IRON   GIRDERS.  CHAP.   XX. 

Here    /=  50,     c   =15,    /=  155    and    k  —  9000,    equals 
the  safe  tensile  power  of  wrought-iron  ;  and  by  (277.) 


30  x  9000 

or  the  depth  should  be  4  feet  ;|  inches.  The  depth  may 
be  found  by  this  formula,  and  then  the  area  of  flanges  by 
formula  (2  74-)  for  assembly  rooms,  banks,  etc. ;  or,  by  for- 
mula (275.}  for  first-class  stores. 


QUESTIONS   FOR   PRACTICE 


574. — In  a  tubular  girder  50  feet  long,  3  feet  4  inches 
high,  and  loaded  with  100,000  pounds  at  the  middle  :  What 
ought  to  be  the  area  of  each  of  the  top  and  bottom  flanges, 
when  the  metal  of  which  they  are  made  may  be  safely 
trusted  with  9000  pounds  per  inch  ? 

575. — In  the  same  girder:  What  should  be  the  area  of  the 
top  or  bottom  flange,  if  the  load  of  100,000  pounds  be  placed 
at  15  feet  from  one  end,  instead  of  at  the  middle  of  the 
beam? 

576. — In  a  tubular  girder  50  feet  long,  40  inches  high, 
and  uniformly  loaded  with  200,000  pounds :  What  should 
be  the  area  of  the  top  and  bottom  flanges,  at  every  five  feet 
of  the  length  of  the  girder? 

577.— In  the  same  girder:  What  ought  to  be  the  thick- 
ness of  the  web,  at  every  five  feet  of  the  length  of  the  girder, 
to  effectually  resist  the  shearing  strain  ? 


QUESTIONS   FOR   PRACTICE.  385 

578. — In  a  tubular  girder  40  feet  long,  32  inches  high, 
sustaining,  with  other  girders  and  the  walls,  the  floor  of  an 
assembly  room,  composed  of  9  inch  70  pound  beams  5 
feet  from  centres,  the  girders  being  placed  16  feet  from 
centres :  What  should  be  the  area  of  each  of  the  top  and 
bottom  flanges,  at  every  five  feet  of  the  length,  the  metal  in 
the  flanges  being  such  as  may  be  safely  trusted  with  9000 
pounds  per  inch  ? 

579. — In  a  floor,  where  the  depth  of  the  tubular  girders 
is  not  arbitrarily  fixed,  where  the  girders  are  42  feet  long 
and  placed  17  feet  from  centres,  and  where  the  total  load 
to  be  carried  is  160  pounds  per  superficial  foot :  What 
would  be  the  proper  depth  of  the  girders,  putting  the  safe 
tensile  strain  upon  the  metal  at  9000  pounds  ? 


CHAPTER    XXL 

CAST-IRON      GIRDERS. 

ART.  580. — Cast-Iron     Superseded    by    Wrouglit-Iron. — 

The  means  for  the  manufacture,  of  rolled-iron  beams  (Chapter 
XIX.)  have  so  multiplied  within  the  last  ten  years  that  the 
cost  of  their  production  has  been  much  reduced,  and  as  a 
consequence  this  beam  is  now  so  extensively  used  as  to 
have  almost  entirely  superseded  the  formerly  much  used 
cast-iron  beam  or  girder.  Beams  and  girders  of  cast-iron, 
however,  are  still  used  in  some  cases,  and  it  is  well  to  know 
the  proper  rules  by  which  to  determine  their  dimensions. 
A  few  pages,  therefore,  will  here  be  devoted  to  this  purpose. 


581.— Flanges— Their    Relative   Proportion. — In   Fig.  75 

we  have  the  usual  form  of  cross-section  of  cast-iron  beams, 
in  which  the  bottom  flange  AB  contains  four  times  as  much 
metal  as  the  top  flange  CD.  It  was 
customary,  fifty  years  since,  to  make 
the  top  and  bottom  flanges  equal.  (See 
Tredgold  on  Cast-Iron,  Vol.  I.,  Art.  37, 
Plate  I.) 

Mr.  Eaton  Hodgkinson  (who  in  1842 
edited  a  fourth  edition  of  Tredgold's 
first  volume,  and  in  1846  added  a 
second  volume  to  that  valuable  work) 
made  many  important  experiments  on 
cast-iron.  Among  the  valuable  deduc- 


FIG.  75. 


tions  resulting  from  these  experiments  was  this  :  that  cast- 


PROPORTIONS   BETWEEN   FLANGES   AND   WEB.  387 

iron  resists  compression  with  about  seven  times  the  force 
that  it  resists  tension  (Vol.  II.,  Art.  34) ;  and  that  the  form  of 
section  of  a  beam  which  will  resist  the  greatest  transverse 
strain,  is  that  in  which  the  bottom  flange  contains  six  times 
as  much  metal  as  the  top  flange  (Vol.  II.,  Art.  138,  page  440). 
If  beams  of  cast-iron  for  buildings  were  required  to  serve  to 
the  full  extent  of  the  power  of  the  metal  to  resist  rupture, 
the  proportion  between  the  areas  of  top  and  bottom  flanges 
should  be  as  i  to  6.  If,  on  the  other  hand,  they  be 
subjected  only  to  very  light  strains,  the  areas  of  the  two 
flanges  ought  to  be  nearly  if  not  quite  equal.  In  view  of  the 
fact  that  in  practice  it  is  usual  to  submit  them  to  strains 
greater  than  the  latter,  and  less  than  the  former,  therefore 
an  average  of  the  proportions  required  in  these  two  cases 
is  that  which  will  give  the  best  form  for  use.  Guided  by 
these  considerations,  it  is  found  that  when  the  flanges  are  as 
i  to  4,  we  have  a  proportion  which  approximates  very 
nearly  the  requirements  of  the  case. 

582. — Flaiige§     ami     Web— Relative    Proportion. — The 

web,  or  vertical  part  which  unites  the  top  and  bottom 
flanges,  needs  only  to  be  thick  enough  to  resist  the  shearing 
strain  upon  the  metal ;  a  comparatively  small  requirement. 
Owing,  however,  to  a  tendency  in  castings  to  fracture  in 
cooling,  the  thickness  of  the  web  should  not  be  much  less 
than  that  of  the  flanges,  and  the  points  of  junction  between 
the  web  and  flanges  should  be  graduated  by  a  small  bracket 
or  easement  in  each  angle.  (Tredgold's  Cast-Iron,  Vol.  II., 
Art.  124.)  The  thickness  of  the  three  parts — web,  top  flange 
and  bottom  flange — may  with  advantage  be  made  in  propor- 
tion as  5,  6  and  8.  Made  in  these  proportions,  the  width  ot 
the  top  flange  will  be  equal  to  one  third  of  that  of  the 
bottom  flange  ;  for  if  wt  equal  the  width  of  the  bottom 
flange  and  wtl  that  of  the  top  flange,  tt  equal  the  thick- 


388  CAST-IRON   GIRDERS.  CHAP.   XXI. 

ness  of  bottom  flange  and    tu   that  of  the  top,  a,   equal  the 
area  of  the  bottom  flange  and    aa    the  area  of  the  top  flange, 

then  a  =  wft   and    wt  —  ~~  ;    also,    aa  =  wata    and    wu  =  —  -  ; 
and    from    these,    remembering    that     at  =  4^,     and     that 

/,  :  t,  :  :  6  :  8, 
we  have    • 


or  the  width  of  the  top  flange  equals  one  third  of  that  of  the 
bottom  flange. 

583.  —  Load  at  middle.  —  Mr.  Hodgkinson  found,  in  his 
experiments,  that  the  strength  was  nearly  as  the  depth  and 
as  the  area  of  the  bottom  flange.  For  the  breaking  weight, 
IV,  he  gives 


an  expression  for  the  relative  values  of  the  dimensions  and 
weight  ;  in  which  W  is  the  breaking  weight  at  the  middle, 
/  the  length  of  the  beam,  d  its  depth,  at  the  area  of  the 
bottom  flange,  and  c  is  a  constant,  to  be  derived  from  ex- 
periment. This  constant,  when  the  weight  was  in  tons  and 
the  dimensions  all  in  inches,  he  found  to  be  26.  Taking  the 
weight  in  pounds  and  the  length  in  feet,  we  have  4853^ 
for  the  constant,  or  say  4850,  and  therefore 

jr=485ogX 

When    a    is  the  factor  of  safety, 

_ 


al 


LOAD  AT   MIDDLE.  389 

which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  cast- 
iron  beam,  required  to  sustain  safely  a  load  at  the  middle. 

The  area  of  the  top  flange  is  to  be  made  equal  to     — ,    and 

4 

the  thicknesses  of  the  web  and  top  and  bottom  flanges  are  to 
be  in  proportion  as    5,    6   and    8. 


584.  —  Example.  —  What  should  be  the  dimensions  of  the 
cross-section  of  a  cast-iron  beam  20  feet  long  between 
supports  and  24  inches  high  at  the  middle,  where  it  is  to 
carry  30,000  pounds  ;  with  the  factor  of  safety  equal  to  5  ? 

Here  W  '  —  30000,  0=5,  /  =  20  and  d=  24  ;  and, 
by  formula  (279.), 

30000  x  5  x  20  ' 

=25'773 


or  the  area  of  the  bottom  flange  should  be  25!  inches. 
Now  the  thickness  will  depend  upon  the  width,  and  this  is 
usually  fixed  by  some  requirement  of  construction.  If  the 
width  be  12  inches,  then  the  thickness  of  the  bottom  flange 

will  be       ^?=2  -IS,     or     2\     inches  full.     The  width  of 

the  top  flange  will  equal  -  =  4     (see  Art,  582),  and 

its  thickness  will  be  •|//  =  |-x2-i5=i.6i,  or  if  inches  ; 
while  the  thickness  of  the  web  will  be  •§-/,  =  |-  x  2  •  1  5  =  I  •  34, 
or  i-  inches. 


585. — Load  Uniformly  Distributed. — A  load  uniformly 
distributed  will  have  an  effect  at  any  point  in  a  beam  equal 
to  that  which  would  be  produced  by  half  of  the  load  if  it 
were  concentrated  at  that  point  (Art.  214).  Therefore,  if 


390  CAST-IRON   GIRDERS.  CHAP.    XXL 


U    equals  the  load  uniformly  distributed,    %U  —  W    in  for- 
mula (879.),  or 

\Ual 


a,  = 


4850^2? 
Ual 


a.  =  - 


(280.) 


which   is  a  rule  for  cast-iron   beams  to  carry  a  uniformly 
distributed  load. 

This  is  precisely  the  same  as  the  previous  rule,  except 
in  the  coefficient.  The  example  given  in  Art.  584  will 
therefore  serve  to  illustrate  this  rule,  as  well  as  the  previous 
one. 


586. — Load    at    Any    Point— Rupture. — From     formula 
(£78.)  we  have 

Wl  =  ca,d 


and,  by  a  comparison  of  formulas  (&1.)  and 


therefore,  in  the  above,  substituting  this  value  of     Wl,     we 
obtain 


(*«•) 


which  is  a  rule  for  the  area  of  the  bottom  flange  at  any  point 
in  the  length  of  the  beam.  The  weight  given  by  this  rule 
is  just  sufficient  to  rupture  the  bottom  flange. 


RULES   FOR   SAFE   LOADS.  39! 

587.  —  Safe  Load  at  Any  Point.—  The  value  of    c,     for  a 
concentrated  load,  is  (Art.  583)     4850,      hence 

4  =      4  i 

C   ~~    4850  ~~    1212^ 

In  formula  (281.),  substituting  for   :      this  value,  and  in- 
serting    a,     the  factor  of  safety,  then 

Wamn 


which  is  a  rule  for  the  area  of  the  bottom   flange  at  any 
point  ;     W,     the  safe  load,  being  concentrated  at  that  point. 

588.  —  Example.  —  In  a  cast-iron  beam  20  feet  long  be- 
tween bearings  :  What  should  be  the  area  of  the  bottom 
flange  at  eight  feet  from  one  end,  at  which  point  the  beam  is 
20  inches  high  and  carries  25,000  pounds;  the  factor  of 
safety  being  equal  to  5  ? 

Here  ^=25000,  0=5,  m  =  S,  n  =  12,  d—2Q  and 
l—2Q\  and  by  formula  (282.) 


25000  x  5  x  8  x  i2 

I2I2|X  20  X  2C 

or  the  area  should  be     24f     inches. 


1212-j-X  20  X  20  24'74 


589. — Safe  Load  Uniformly  I>i§tributed— Effe<*  at  Any 
Point. — This  effect  at  any  point  is  equal  to  that  produced  by 
half  the  load  were  it  all  concentrated  at  that  point  (Art. 
214);  therefore,  if  U  represent  the  uniformly  distributed 
load,  then  by  formula  (282.) 


392  CAST-IRON   GIRDERS.  CHAP.   XXI. 

which  is  a  rule  for  the  area  of  the  bottom  flange  of  a  cast- 
iron  beam  at  any  point,  to  carry  safely  a  uniformly  dis- 
tributed load.  If  the  depth  of  the  beam  remain  constant 
throughout  the  length*  then  at  will  vary  as  the  rectan- 
gle mn. 

From  formula  (283.)  we  have 


which  is  a  rule  for  the  depth  of  a  beam  at  any  point,  to  carry 
safely  a  uniformly  distributed  load.  If  the  area  of  the 
bottom  flange  remain  constant  throughout  the  length,  then 
d  will  vary  as  the  rectangle  mn. 

590.— Form  of  Web.— By  the  last  formula,  (284.),  it  will 
be  seen  that  when  at,  '  the  area  of  the  bottom  flange,  re- 
mains constant  throughout  the  length  of  the  beam,  then  the 
depths  will  vary  in  proportion  to  the  rectangle  of  the  two 
segments,  m  and  ;/,  of  the  length.  The  corresponding 
curve  which  may  be  drawn  through  the  tops  of  the  ordinates 
denoting  the  various  depths,  is  that  of  a  parabola  (Art.  212). 
Instead  of  computing  the  depths  at  frequent  intervals,  there- 
fore, it  will  be  sufficient  to  compute  the  depth  at  the  centre 
only,  and  then  give  to  the  web  the  form  of  a  parabola. 


591.  —  Two  Concentrated  Weight§—  Safe  Load.  —  Formula 
is  appropriate  for  a  concentrated  load  at  any  point  in 
the  length  of  a  beam,  and  formula  (30.)  is  for  two  concen- 
trated loads  at  any  given  points. 

A  comparison  of  these  formulas  shows  that 


~ 


LOADED   WITH   TWO   WEIGHTS.  393 

In  Art.  586  we  have 


which  is  an  expression  for  the  breaking  load.     Inserting     a, 
the  symbol  of  safety,  in  this  expression,  we  have 


mn 


catd  —  4  Wa  —, 

an  expression  for  the  safe  load  for  cast-iron  beams.  If  for 
the  second  member  of  this  equation  there  be  substituted  its 
value  as  above, 


we  shall  have 

ca/d  =  4a™-(Wn+Vs) 

an  expression  for  two  concentrated   safe  loads.     From  this 
we  have 

A  1  1  J 

Vs) 


In  Art.  587  we  have    —  = =,     therefore 

c        1212%' 


m or 


td  =  a  -,-  (  Wn  +  Vs) 


m 


a-(Wn+Vs) 


which,  in  a  beam  carrying  two  concentrated  loads,  is  a  rule 


394 


CAST-IRON    GIRDERS. 


CHAP.    XXL 


for  the  area  of  the  bottom  flange  at  the  location  of 
of  the  loads,  as  in  Fig.  76;  and  (see  Art.  153) 


one 


Wm) 


(286.) 


which,  in  a  beam  carrying  two  concentrated  loads,  is  a  rule 
for  the  area  of  the  bottom  flange  at  the  location  of  F,  one 
of  the  loads,  as  in  Fig.  76. 


592. — Examples. — As  an  application  of  rules  (285.)  and 
(286),  let  it  be  required  to  ascertain  the  dimensions  of  a  cast- 
iron  girder  to  sustain  a 
brick  wall  in  which  there 
are  three  windows,  as  in 
Fig.  76,  so  disposed  as  to 
concentrate  the  weight  of 
the  wall  into  two  loads,  as 
at  W  and  V.  Let  /, 
the  length  in  the  clear  of 
the  supports,  =  20,  m  —  7, 
n  =  13,  s  =  6  and  r  =  14 
feet,  and  the  height  of  the  girder  at  W  and  V  equal  25 
inches.  Also,  let  the  wall  be  16  inches  thick,  and  so  much 
of  it  as  is  sustained  at  W  measure  250  cubic  feet,  at  no 
pounds  per  foot,  or  27,500  pounds.  Likewise,  suppose  the 
weight  upon  V  to  equal  27,000  pounds. 

Taking   the   factor   of  safety   at     5     we   now   have,   by 
formula  (285. \ 


FIG.  76. 


a  —  5  x  A  K275QQ  x  13)  +  (27000  x  6)]  _ 

X  25 


29-99 


SUSTAINING   TWO    BRICK    PIERS.  395 

or  the  area  of  flange  is  required  to  be    30    inches  at     W\ 
and,  by  formula  (286.), 

5  x  A  [(27000  x  14)  +  (27500  x  7)] 
a,  •=.  --  1  --  -  --  =  2o-23 
'  1212^x25 

or  the  area  of  flange  is  required  to  be    28^    inches  at     V. 

As  the  wall  is  16  inches  thick,  the  width  of  the  bottom 
flange  should  be  16  inches,  and  its  thickness  therefore 
should  be 

^  =  1-875     inches  at     W 

?  8  •  ?  i 

ft  .  —  i  •  764     inches  at     V 

From  W  to  V  the  thickness  is  to  be  graded  regularly 
from  1-875  to  1-764;  while  from  W  to  the  end  'next 
W  it  is  to  be  equal  to  that  at  W,  i-J-  inches  thick,  and 
from  V  to  the  end  next  V  it  is  to  be  if  inches  thick. 
The  width  of  the  top  flange  is  to  be  (Art.  582)  one  third 
of  the  width  of  the  bottom  flange,  or  ig-  =  5-J-  inches.  Pro- 
portioning the  three  parts  as  5,  6  and  8  (Art.  582),  the 
thickness  of  the  top  flange  will  be 


-|  x  if  =  i^     inches  at     V 
I  x  i-J=  i-J-f     inches  at     W 

The  thickness  is  to  be  graded  regularly  between  W  and  V, 
and  thence  to  each  end  of  the  beam  the  thickness  is  to  be 
that  of  W  and  V  respectively.  The  web  is  to  be  of  the 
shape  shown  in  Fig.  76,  and  is  to  be  (Art.  582) 

m  \  x  if  =  i^\-    at     V  and 

|xi|=iii     at     W 

or,  say     i£     inches,  averaging  it  throughout. 


396 


CAST-IRON   GIRDERS. 


CHAP.    XXI. 


593. — Arclied  Girder. — A  beam  such  as  shown  in  Fig.  77 
is  known  as  the  "  bow-string  girder,"  in  which   the  curved 

part  is  a  cast-iron  beam  of 
the  T  form  of  cross-section, 
and  the  feet  of  the  arch 
are  held  horizontally  by  a 
wrought-iron  tie-rod.  This 
beam,  although  very  pop- 
FIG.  77.  ular  with  builders,  is  by  no 

means  worthy  of  the  confidence  which  is  placed  in  it. 
With  an  appearance  of  strength,  it  is  in  reality  one  of  the 
weakest  beams  used.  Without  the  tie-rod  its  strength  is 
very  small,  much  smaller  than  if  the  T  section  were  reversed 
so  as  to  have  the  flange  at  the  bottom,  thus,  J.  (Tredgold, 
Vol.  II. ,  pp.  414  and  415). 


594. — Tie-Rod  of  Arched  Girder. — The  action  of  a  con- 
centrated weight  at  the  middle  of  a  tubular  girder,  in 
producing  tension  in  the  bottom  flange,  is  explained  in 
Art.  551.  The  tension  in  the  tie-rod  of  an  arched  girder  is 
produced  in  precisely  the  same  manner,  and  therefore  the 
rule  {form.  265.)  there  given  will  be  applicable  to  this  case, 
when  modified  as  required  for  a  uniformly  distributed  load ; 
or,  for  W  substituting  its  value,  \U  (Art.SQS).  Then,  upon 
the  presumption  that  there  is  sufficient  material  in  the  cast 
arch  to  resist  the  thrust,  we  have 


in  which    d   is  in  feet.     If    d    be  taken  in  inches,  then 

Ul  (287.) 


TIE-ROD    OF  ARCHED    GIRDER.  397 

which  is  a  rule  for  the  area  of  the  cross-section  of  the  tie- 
rod  in  an  arched  girder ;  in  which  at  is  the  area  of  the 
cross-section  of  the  rod,  U  is  the  weight  in  pounds  equally 
distributed  over  the  beam,  /  is  the  length  in  feet  between 
the  supports,  d,  in  inches,  is  the  depth  or  versed  sine  of  the 
arc,  or  the  vertical  distance  at  the  middle  of  the  beam  from 
the  axis  of  the  tie-rod  to  the  centre  of  gravity  of  the  cross- 
section  of  the  cast-iron  arch,  and  k  v  is  the  weight  in  pounds 
which  may  safely  be  trusted  when  suspended  from  the  end 
of  a  vertical  rod  of  wrought-iron  of  one  square  inch  section. 
If  this  latter  be  put  at  9000  pounds,  then 

Ul 


Now    at    is  the  area  of  the  tie-rod.     The  area  of  any  circle 
is  equal  to  the  square  of  its  diameter  multiplied  by   -7854,  or 

«,==  -7854^ 
and  since,  by  formula  (287. \ 

therefore 
and 


If  k,   the  safe  resistance  to  tension  per  inch,  be  taken  at 
9000-  pounds,  the  rule  becomes 


which  is  a  rule  for  the  .diameter  of  the  tie-rod  of  an  arched 
girder. 


398 


CAST-IRON    GIRDERS. 


CHAP.    XXI. 


595. — Example.  — What  should  be  the  diameter  of  the 
tie-rod  of  an  arched  girder,  20  feet  long  in  the  clear  between 
supports,  and  24  inches  high  from  the  axis  of  the  tie-rod  to 
the  centre  of  gravity  of  the  cross-section  at  the  middle  of  the 
arched  beam  ;  the  load  being  40,000  pounds  equally  dis- 
tributed over  the  length  of  the  beam  ? 

Here  we  have  U  =  40000,  /  —  20  and  ^=24;  and 
therefore,  by  formula  (289.\ 


4712  x  24 


or  the  diameter  of  the  rod,  with  the  safe  resistance  to  tension 
taken  at  9000   pounds,  should  be   2f   inches. 


596. — Substitute  for  Arched  Girder. — The  cast-iron  arch 
of  an  arched  girder  serves  to  resist  compression.  Its  place 
can  as  well  be  filled  by  an  arch  of  brick,  footed  on  a  pair  of 
cast-iron  skew-backs,  and  these  held  in  position  by  a  pair  of 
tie-rods,  as  in  Fig.  78. 


FIG.  78. 

To  obtain  a  rule  for  the  diameter  of  each  rod,  we  have  as 
above,  in  Art.  594, 

*,=  -7854^ 


BRICK   SUBSTITUTE   FOR   IRON   ARCH.  399 

This  is  for  one  rod.     When   at  is  put  for  the  joint  area  of  two 
rods,  we  will  have 


Comparing  this  with  formula  (287.  \  we  have 

til 


or 


f  X2  X   •  78540$ 

and  when  k   is  taken  at  9000  (Art.  594) 

Ul 


(290.) 
9425^ 

This  is  a  rule  in  which  Df  represents  the  diameter  of  each 
of  the  two  required  rods. 

For  example,  see  Art.  595. 

An  arch  of  brick,  well  laid  and  secured  in  this  manner, 
will  serve  quite  as  well  as  the  cast-iron  arch,  and  may  be 
had  at  less  cost.  The  best  supports,  however,  to  carry  brick 
walls  are  those  made  of  rolled-iron  beams,  putting  two  or 
more  of  them  side  by  side  and  bolting  them  together.  (See 
Art.  489,  form. 


400  CAST-IRON   GIRDERS.  CHAP.   XXI. 


QUESTIONS   FOR   PRACTICE. 


597. — What  should  be  the  dimensions  of  cross-section  of 
a  cast-iron  girder,  23  feet  long  between  supports,  and 
27  inches  high  at  the  middle,  at  which  point  it  is  to  carry 
40,000  pounds  ;  with  5  as  the  factor  of  safety  ?  The  width 
of  bottom  flange  is  16  inches. 

598. — In  a  girder  of  the  same  length,  height  and  width  : 
What  should  be  the  cross-section  if  the  weight  be  60,000 
pounds  and  be  uniformly  distributed  ;  the  factor  of  safety 
being  5  ? 

599. — In  a  girder  of  the  same  length,  and  of  the  same 
height  and  width  at  8  feet  from  one  end,  where  it  is  required 
to  carry  50,000  pounds,  with  a  factor  of  safety  of  5  :  What 
should  be  the  dimensions  of  cross-section  ? 

600. — In  a  girder  25  feet  long  between  bearings,  carry- 
ing a  load  of  40,000  pounds  at  10  feet  from  one  end,  with 
5  as  the  factor  of  safety,  and  having  30  inches  area  of  cross- 
section  in  the  bottom  flange :  What  should  be  the  depth  of 
the  girder  ? 

60 1. — A  girder,  25  feet  long  and  30  inches  high,  is  re- 
quired to  carry,  with  5  as  a  factor  of  safety,  two  weights, 
one  of  25,000  pounds  at  8  feet  from  one  end,  and  the  other 
of  30,000  pounds  at  6  feet  from  the  other  end  :  What  should 


QUESTIONS   FOR   PRACTICE.  4OI 

be  the  dimensions  of  cross-section  at  each  weight,  the  bottom 
flange  being    16   inches  wide? 

602. — In  an  arched  girder,  24  feet  long  between  bear- 
ings, with  a  versed  sine  or  height  of  30  inches  from  the  axis 
of  the  rod  to  the  centre  of  gravity  of  the  arched  beam  at 
the  middle,  and  with  the  load  on  the  girder  taken  at  80,000 
pounds  uniformly  distributed :  What  ought  the  diameter  of 
the  tie-rod  to  be  ? 


CHAPTER    XXII. 

FRAMED    GIRDERS. 

ART.  603. — Transverse  Strains  in  Framed  Girders. — This 
work,  a  treatise  elucidating  the  Transverse  Strain,  would 
seem  to  have  reached  completion  with  the  end  of  the  discus- 
sion on  simple  beams  ;  but  when  it  is  recognized  that  the 
formation  of  a  deep  girder,  by  a  combination  of  various 
pieces  of  material,  is  but  a  continuation  of  the  effort  to  gain 
strength  in  a  beam,  by  concentrating  its  material  far  above 
and  below  the  neutral  axis,  as  is  done  in  the  tubular  girder 
and  rolled-iron  beam,  it  is  clear  that  the  subject  of  framed 
girders  is  properly  included  within  a  treatise  upon  the 
transverse  strain.  The  subject  of  framed  girders,  however, 
will  here  be  discussed  so  far  as  to  develop  only  the  more  im- 
portant principles  involved.  For  examples  in  greater  vari- 
ety, the  reader  is  referred  to  other  works  (Merrill's  Iron 
Truss  Bridges,  and  Bow's  Economics  of  Construction). 


604. — Device  for  Increasing  the  Strength  of  a  Beam.— 

The  use  of  simple  beams  is  limited  to  comparatively  short 
spans  ;  for  beams  cut  from  even  the  largest  trees  can  have 
but  comparatively  small  depth.  The  po\ver  of  a  beam  to 
resist  cross-strain  can  be  considerably  increased  by  a  very 
simple  device.  Let  Fig.  79  represent  the  side  view  of  a  long 
beam  of  wood,  from  which  let  ACDB,  the  upper  part  of  the 
beam,  be  cut.  With  the  pieces  thus  removed,  and  the  addi- 
tion of  another  small  piece  of  timber,  there  may  be  con- 


INCREASING  STRENGTH   OF   BEAM. 


403 


structed  the  frame  shown  in  Fig.  80,  which  is  capable  of  sus- 
taining a  greatly  increased  load.     This  increase  will  be  in 


FIG.  79. 


FIG.  80, 


proportion  to  the  depth  of  the  frame  (Art.  583),  and  is  ob- 
tained here  by  increasing  the  distance  between  the  fibres 
which  resist  compression  and  those  which  resist  tension.  It 
is  upon  this  principle  that  roof  trusses  and  bridge  girders, 
alike  with  common  beams,  all  depend  for  their  stability. 


605. — Horizontal  Thrust.— In  a  frame  such  as  Fig.  Bo, 
the  horizontal  strains  produced  by  the  weight  W  are  bal- 
anced ;  or,  the  tension  caused  in  the  tie  CD  is  equal  to  the 
compression  caused  in  the  short  timber  on  which  the  weight 
rests.  If  the  tie  CD  were  removed,  it  is  obvious  that  the 
weight  W,  acting  through  the  two  struts  AW  and  BW, 
would  push  the  two  abutments  AC  and  BD  from  each 
other,  and,  descending,  fall  through  between  them  ;  unless 
the  abutments  were  held  in  place  by  resistance  other  than 
that  contained  in  the  frame — such,  for  instance,  as  outside 
buttresses. 


404 


FRAMED    GIRDERS. 


CHAP.   XXII. 


From  this  we  learn  the  importance  of  a  tie-beam ;  or,  in 
its  absence,  of  sufficient  buttresses.  From  this  we  may  also 
learn  why  it  is  that  roof  trusses  framed  without  a  horizontal 
tie  at  foot  so  invariably  push  out  the  walls,  when  constructed 
without  exterior  buttresses. 


606. — Parallelogram  of   Forces— Triangle  of  Forces. — 

A  discussion  of  the  subject  of  framed  girders  can  only  be  in- 
telligently understood  by  those  who  are  familiar  with  some 
of  the  more  simple  and  fundamental  principles  of  statics. 
One  of  these  principles  is  known  as  the  parallelogram  of 
forces,  or  the  triangle  of  forces,  and  is  useful  to  the  archi- 
tect in  measuring  oblique  strains  due  to  vertical  and  hori- 
zontal pressures.  Proof  of  the  truth  of  this  principle  may 
be  found  in  most  mathematical  works.  (See  Cape's  Math., 
Vol.  II.,  p.  118  ;  chap,  on  Mechs.,  Art.  20.)  In  this  chapter 
its  application  to  construction  will  be  shown. 

In  Fig.  81,  let  the  lines   A  W   and    BW  represent  the  axes 
of  two  timber  struts,  which,  meeting  at  the  point     Wt    sus- 


FIG.  81. 

tain  a  weight,  or  vertical  pressure,  as  indicated  by  the  arrow 
at     W.      Then,    let   the  vertical   line     WE,    drawn    by  any 


PARALLELOGRAM  OF  FORCES.  405 

convenient  scale,  represent  the  number  of  pounds,  or  tons, 
contained  in  the  vertical  weight  at  W.  From  E,  draw 
ED  parallel  with  A  W,  and  EC  parallel  with  BW. 
CWDE  is  the  parallelogram  of  forces,  and  possesses  this 
important  property  —  namely,  that  the  three  lines  WE,  EC 
and  CW,  forming  a  triangle,  are  in  proportion  to  three  forces  ; 
the  weight  at  W,  the  strain  in  WB,  and  the  strain  in  WA. 
The  same  is  true  of  the  other  triangle  WED  ;  or,  to  des- 
ignate more  particularly,  we  have  :  —  as  the  line  WE  is  to 
the  weight  at  W,  so  is  the  line  CE,  or  WD,  to  the  strain 
in  WB;  and  also  :—  as  the  line  WE  is  to  the  weight  at 
W,  so  is  the  line  DE,  or  WC,  to  the  strain  in  A  W.  In- 
dicating the  lines  by  the  letters  a,  b  and  ct  as  in  the  fig 
ure,  we  have 


c  :  a  :  :  W/  :  A, 


in  which    At    equals  the   strain   caused   by  the  weight     Wt 
through  the  line     WA  ;     and 


c  :  b  :  :    W,  :  Bt 
B,  =-  W~ 


in   which    Bt    equals  the   strain   caused   by  the  weight     Wf 
through  the  line     WB. 


60  7  B — Line§  and  Force§  in  Proportion. — The  above  pro- 
portions hold  good  when  the  two  lines  A  W  and  BW  are 
inclined  at  any  angle,  and  whether  they  are  of  equal  or  of  un- 
equal lengths  ;  indeed,  the  principle  is  general  in  its  applica- 


406 


FRAMED    GIRDERS. 


CHAP.   XXII. 


tion,  for  in  all  cases  where  the  three  sides  of  a  triangle  are 
respectively  drawn  parallel  to  the  direction  of  three  several 
forces  which  are  in  equilibrium,  then  the  lengths  of  the  three 
lines  will  be  respectively  in  proportion  to  the  three  forces. 


608. — Horizontal  Strain  Measured  Graphically. — In  Fig. 

81,  and  in  the  triangle     WCE,    draw,    from    C,    the  horizon- 
tal line    CF,   or    h ;    then  we  have  the  line    £,    in  proportion 


FIG.  81. 

to  the   line    //,    as    Bn    the  strain   in     WB,    is  to    Ht,    the 
horizontal  strain  ;  or, 

b  :  h  ::  B,  :  ff,-B^ 
and  by  substituting  the  value  of   Bt    in  formula  (292.)  have 


H,  =:  Bj-  =  W~  =  W£ 
'          <b  'cb  'c 


or  the  horizontal  strain  is  measured  by  the  quotient  arising 
from  a  division  by  the  line    c,    of  the  product  of  the  weight 


FORCES   IN   EQUILIBRIUM, 

Wt    into  the  line    h  ;     or, 

c  :  h  : :   W,  :  H, 


407 


(293.) 


This  measures  the  horizontal  strain  at  AB,  or  at  W,  for 
it  is  the  same  at  all  points  of  such  a  frame,  whatever  the 
angle  of  inclination  of  the  struts,  or  whether  they  are  inclined 
at  equal  or  unequal  angles. 


609. — Measure  of  Any  Number  of  Forces  in  Equilibrium. 

— In  Fig.  82,  let  AB  be  the  axis  of  a  horizontal  timber,  sup- 
ported at  A  and  B,  and  let  AC,  CD  and  DB  be  three 
iron  rods,  with  two  weights  R  and  P  suspended  from  the 


FIG.  82. 

points  C  and  D.  The  iron  rods  being  jointed  at  A,  C, 
D  and  B,  so  as  to  permit  the  weights  to  move  freely,  and 
thus  to  adjust  themselves  to  an  equilibrium,  the  whole  frame 
ABDC  will  be  equilibrated. 

From  D  erect  a  vertical,  DH,  and  by  any  convenient 
scale  make  DG  equal  to  the  weight  P,  and  GH  equal  to 
the  weight  R.  From  £,  draw  GE  parallel  with  CD,  and 
from  H  draw  HE  parallel  with  AC.  The  sides  of  the 


408 


FRAMED    GIRDERS. 


CHAP.    XXII, 


triangle  GED  are  parallel  with  the  three  lines  CD,  DB 
and  DP,  and  consequently  are  in  proportion  as  the  strains 
in  the  three  lines  CD,  DB  and  DP.  Again ;  the  sides  of 
the  triangle  HEG  are  parallel  with  the  lines  AC,  CD  and 
CR,  and  consequently  are  in  proportion  as  the  strains  in  the 
lines  AC,  CD  and  CR.  From  E  draw  EF  horizontal. 
Then  the  sides  of  the  triangle  FED,  being  parallel  with  the 
lines  BA,  BD  and  BK,  are  in  proportion  to  the  strains  in 
these  lines.  Also,  the  sides  of  the  triangle  HEF,  being 
parallel  to  the  lines  AB,  AC  and  AL,  are  in  proportion 


FIG.  82. 

to  the  strains  in  these  lines.  Thus,  in  the  triangles  within 
HDE,  we  have  the  measures  of  all  the  strains  of  the  funicu- 
lar or  string  polygon  ABDCA  ;  FE  being  the  horizontal 
strain,  FD  the  vertical  strain  or  load  on  BK,  and  HF 
the  vertical  strain  or  load  on  AL. 


610. — Strain*  in  ait  Equilibrated  Truss. —  In  Fig.  82  the 
strains  in  the  lines  AC,  CD  and  DB  are  tensile,  while  that 
in  AB  is  compressive.  If  the  lines  AC,  CD  and  DB  were 
above  the  line  AB,  instead  of  below  it,  then  these  strains 
would  all  be  reversed ;  those  which  are  tensile  in  the  figure 
would  then  be  compressive,  while  that  which  is  com- 


EQUILIBRATED   FRAME. 


409 


pressive  would  then  be  tensile  ;  but  the  amount  of  strain  in 
each  would  be  the  same  and  be  measured  as  in  Fig.  82. 

For  example  :  Let  Fig.  83  represent  an  equilibrated  frame  ; 
the  pieces  A  C,  CD,  DE,  EF  and  FB  suffering  compression 


FIG.  83. 


from  the  vertical  pressures  indicated  by  the  arrows  at  C,  D, 
E  and  F,  while  AB,  a  tie,  prevents  the  frame  from  spread- 
ing. Draw  the  vertical  line  LQ,  and  from  B  draw  radiating 
lines,  parallel  respectively  with  the  several  lines  AC,  CD, 
DE  and  EF,  and  cutting  the  line  LQ  at  the  points  Q,  P, 
O  and  M.  Then  the  several  lines  BQ,  BP,  BO,  BM  and 
BL  will  be  in  proportion,  respectively,  to  the  strains  in 
AC,  CD,  DE,  EF  and  FB  ;  and  the  lines  LM,  MO,  OP 
and  PQ  will  be  in  proportion,  respectively,  to  the -vertical 
pressures  at  F,  E,  D  and  C ;  while  the  line  LN  will 
represent  the  vertical  pressure  on  B,  and  NQ  that  on  A, 
and  the  line  NB  the  horizontal  thrust  in  AB. 


611. — From    Given  Weigh!*    to    Construct    a    Scale    of 

Strains. — The  construction  of  the  scale  of  strains   LBQ,    as 
here  given,  is  proper  in  a  case  where  the  points    C,  D,  E 


4io 


FRAMED   GIRDERS. 


CHAP.    XXII. 


and  F  are  fixed,  and  the  weights  and  strains  are  required. 
When  the  weights  at  C,  D,  E  and  F,  with  their  horizontal 
distances  apart,  and  the  two  heights  RC  and  UF,  are 
given  ;  then,  to  find  the  scale  of  strains,  and  incidentally  the 
heights  of  the  points  D  and  E,  proceed  thus :  From  B 


FIG.  83. 


draw  BQ  parallel  with  AC,  make  the  vertical  QL  equal, 
by  any  convenient  scale,  to  the  sum  of  the  weights  at  C,  D, 
E  and  F,  and  upon  this  vertical  lay  off  in  succession  the 
distances  LM,  MO,  OP  and  PQ,  equal  respectively  to 
the  weights  at  F,  E,  D  and  C.  Then,  the  several  lines 
BL,  BM,  BO,  BP  and  BQ  will,  by  the  same  scale, 
measure  the  several  strains  in  BF,  FE,  ED,  DC  and  CA, 
and  BN  will  measure  the  horizontal  strain. 


612. — Example. — In  constructing  Fig.  83,  the  weights 
given  are  11,899  pounds  at  F,  4253  pounds  at  E,  4464 
at  D  and  11,384  at  C\  being  a  total  of  32,000  pounds. 
The  distances  AR,  RS,  etc.,  are  successively  8,  13,  20, 
24  and  13;  in  all  78  feet.  The  height  RC  =  15,  and 
UF  = 


CONSTRUCTING  A   SCALE   OF   STRAINS.  411 

With  these  dimensions  all  laid  down  as  in  Fig.  83,  draw 
BQ  parallel  with  AC.  Draw  the  line  LQ  vertical,  and  at 
such  a  distance  from  B  as  that  its  length  shall,  by  a  scale 
of  equal  parts,  be  equal  to  the  total  load  on  the  four  points 
C,  D,  E  and  F ;  or  to  a  multiple  of  the  total  load.  For 
example  :  a  scale  of  100  parts  to  the  inch  will  be  convenient 
ki  this  case,  by  appropriating  4  parts  to  the  thousand 
pounds.  The  32,000  pounds  require  32x4=128  parts  for 
the  length  ol  the  line  LQ,  and  the  several  other  weights 
and  distances  require  as  follows : 

LM  —  4  x  1 1  •  899  =  47  -  596 

MO  —  4  x    4-253  —  17-012 

OP  —  4  x    4-464=  17-856 

PQ  =  4x  11-384  ^  45.536 

The  sum  of  these, 

LQ  =  47-596  4-  17-012  +'17-856  +  45-S36  =  128 

as  before.  Therefore,  draw  LQ  at  such  a  distance  from  B 
that  it  will,  by  the  scale  named,  equal  128  parts.  On  this 
line  lay  off  the  distances  LM  =  47-596,  MO  =  17-012,  etc., 
as  above  given.  Join  B  with  each  of  the  points  P,  O  and 
M.  These  lines  give  the  directions  of  the  lines  CD,  DE 
and  EF ;  therefore,  draw  FE  parallel  with  BM,  ED 
parallel  with  BO,  and  DC  parallel  with  BP. 

By  applying  the  scale  to  the  lines  radiating  from  B,  the 
strains  in  the  several  lines  AC,  CD,  etc.,  will  be  shown. 

BQ,  by  the  scale,  measures  80  parts,  therefore  \°-  =  20  ; 
or  the  strain  in  A  C  is  20,000  pounds. 

BP  measures  45  parts,  and  -\5  =  n£;  or  the  strain  in 
CD  is  11,250  pounds. 

BO  measures  38,  and  $•£-  =  9^ ;  or  the  strain  in 
DE  is  9500. 

BM  measures  39,  and  -3T9-  =  9!  ;  or  the  strain  in 
EF  is  9750. 


412  FRAMED    GIRDERS.  CHAP.    XXII. 

(5o  •  ^ 
BL    measures    69.5,   and     ——=17-375;    or  the  strain 

in   FB    is    17,375. 

BN  measures  37-5.  and  -- —  =  9.375;  or  the  horizon- 
tal strain  is  9375  pounds. 

Also,  as  LN  measures  58,  therefore  ^-  =  14,500,  equals 
the  load  on  B  \  and  as  NQ  measures  70,  therefore, 
-\°-  =  17,500,  equals  the  load  on  A  ;  and  the  two  loads  At 
and  B,  together  equal  17500  +  14500  =  32000,  equals  the 
total  load. 

In  practice  the  diagram  should  be  large,  for  the  accuracy 
of  the  results  will  be  in  proportion  to  the  size  of  the  scale, 
as  well  as  to  the  care  with  which  it  is  drawn  and  measured. 
The  size  above  taken  is  large  enough  for  the  purposes  of 
illustration  merely,  but  in  practice  the  diagram  should  be 
drawn  at  a  scale  of  12  feet  to  the  inch ;  or,  still  better,  at  8 
feet.  (See  Art.  615.) 

613. — Horizontal    Strain   Measured   Arithmetically. — In 

the  last  article,  directions  were  given  for  locating  the  line 
LQ,  Fig.  83.  This  line  may  be  located  more  precisely  by 
arithmetical  computation,  and  the  horizontal  thrust  be  thus 
denned  more  accurately  than  is  there  done.  In  Fig.  84, 
showing  parts  of  Fig.  83  enlarged,  we  have  the  triangles 
ACR  and  BFU,  the  same  as  in  Fig.  83. 

The  triangle  ACR,  as  stated  (Art.  612),  has  a  base  of  8 
and  a  height  of  15.  Make  A  Y  equal  10,  and  draw  YZ 
vertical.  We  now  have  this  proportion, 

AR  :  RC  : :  A  Y :  YZ  or 

8    :   15   ::    10  :  YZ  =  -°-^  =  18-75 

o 

Again  ;  triangle  BFU,  as  stated  (Art.  612),  has  a  base  of  13 
and  a  height  of  2oJ.  Make  BX  equal  10,  and  draw  XV 


MEASURING   HORIZONTAL   STRAIN.  413 

vertical.     We  have  now  this  proportion, 

BU\  UF \\BX\XV  or 

10  x  2oJ 
13  :2oi::   10  :  XV  = — ±  =  15-577 

Thus  we  have  the  two  angles  at  A  and  B  comparable, 
for,  with  a  common  base  of  10,  the  one  at  A  has  a  height 
of  18-75,  while  the  one  at  B  is  15-577.  The  two  triangles 
may  now  be  put  together  at  the  line  BU.  Extend  the  verti- 
cal line  VX  to  W,  make  XW  equal  to  YZ  =  18-75,  and 
join  W  and  B.  Then  the  triangle  BXW  equals  the 
triangle  A  YZ,  and  BW  is  parallel  with  AZ. 


FIG.  84. 

The  problem  now  is  to  locate  the  point  N,  so  that  the 
vertical  LQ  drawn  through  N  shall  be  equal,  by  any 
given  scale,  to  the  total  of  the  loads  at  C,  D,  E  and  F. 
To  do  this  we  have 


FRAMED    GIRDERS.  CHAP.    XXII, 

BX  x  NL 


BN:NL::BX:XV  = 
BN:NQ::BX:XW  = 


BN 

BXx  NQ 
BN 


By  addition  we  have 

_  BXxNL     BXxNQ  _  BX(NL  +  NQ) 


By  the  diagram,     XV+XW=  VW,     and     NL  +  NQ  =  LQ, 
and  therefore 


.  BN 
VW:  BX-.:  LQ:  BN  = 


or 


BX^LQ 

VW  ~ 


The  total  load  is  32,000,  for  which  we  may,  putting  one  for 
a  thousand,  make  LQ  equal  to  32 ;  and,  since  VW  equals 
YZ+  VX  =  18-75  +  15-577  =  34-327.  and  BX  =  10,  we 
have 


PRESSURE   ON   THE   TWO    POINTS   OF   SUPPORT.  415 


defining  accurately  the   horizontal  thrust     BN    as     9-322, 
or     9322     pounds. 


Vertical     Pressure     upon     the    Two    Points    of 

Support. — This  pressure  was  shown  in  Art.  612  by  the 
diagram,  but  may  be  more  accurately  determined  by  arith- 
"metical  computation,  as  follows :  In  the  last  article  the 
horizontal  thrust  BN  was  shown  to  be  9322  pounds.  We 
have  the  proportion 

BX\     XV    ::    BN  :  NL 

15-577  x  9-322 
10  :  15-  577  1:9-322  :NL  =  -        ^--^       -  =  14-521 

or  the  vertical  strain  upon  the  support  B  is  14,521  pounds. 
To  find  that  upon  the  support  A  we  have 


BX :   XW  ::    BN    :  NQ 


18-75x9.322 


10  :  18-75  ::  9-322  :  NQ  =  -  -  =  17.479 


10 


or  the  vertical  strain  upon  the  support   A    is    17,479  pounds 
and  the  two,     17479  +  14521  =  32000,     equals  the  total  load. 


615. — Strains  measured  Arithmetically. — The  resulting 
strains  in  Fig.  83,  as  obtained  by  scale  in  Art.  6(2,  are  close 
approximations,  and  are  near  enough  for  general  purposes. 
The  exact  results  can  be  had  arithmetically,  as  in  Arts.  613 
and  614.  For  example  :  In  Art.  612  the  horizontal  thrust 
was  found  by  scale  to  be  9375,  while  in  Art.  613  it  was 
more  accurately  defined  by  arithmetical  process  to  be  9322. 
So,  also,  the  portions  of  the  total  load  borne  by  the  two 


41 5  FRAMED    GIRDERS.  CHAP.   XXII. 

supports,  A  and  B,  were  found  by  scale  to  be  17,500 
and  14,500,  respectively,  while  in  Art.  614  they  were  accu- 
rately fixed  at  17,479  on  A  and  14,521  on  B.  A  carefully 
drawn  diagram  at  a  large  scale  will  generally  be  sufficient 
for  use,  but  it  is  well,  in  important  cases,  to  compute  the 
dimensions  also.  When  both  are  done,  the  scale  measure- 
ments serve  as  a  check  against  any  gross  errors  in  the  com- 
putations. 

In  Art.  612  the  strains  in  the  several  timbers  are  given, 
as  ascertained  by  scale.  By  the  rules  for  computing  the 
sides  of  a  right-angled  triangle  (the  47th  of  first  book  of  Eu- 
clid), the  several  strains,  as  represented  by  BL,  BM,  BO, 
etc.  (Fig.  83),  may  be  found  arithmetically.  The  following  list 
shows  the  results  by  this  method,  side  by  side  with  those  by 
scale : 

By  Scale.  By  Computation. 

BL  =  17,375  and  17,256 

BM —    9,750       "  9,684 

BO  —    9,500       "  9,464 

BP  —  11,250       •'  11,138 

BQ  --•  20,000       "  19,810 


616. — Curve     of    Equilibrium— Stable    and     Unstable. — 

When  the  positions  of  the  supporting  timbers  AC,  CD,  DE, 
etc.  (Fig.  83),  are  regulated  in  accordance  with  the  weights 
upon  the  points  C,  D,  E,  etc.,  and,  as  shown  in  Art.  611, 
the  frame  is  in  a  state  of  equilibrium  ;  and  a  curve  drawn 
through  the  points  A,  B,  C,  Dy  etc.,  is  called  the  curve  of 
equilibrium.  When  the  several  weights  are  numerous,  are 
equal,  and  are  located  at  equal  distances  apar.t ;  or,  when 
the  load  is  uniformly  distributed  over  the  length  of  the 
frame,  this  curve  is  a  parabola.  In  these  cases,  if  the  rise 


CURVE   OF   EQUILIBRIUM.  417 

be  small  in  comparison  with  the  base,  the  curve  is  nearly 
the  same  as  a  segment  of  a  circle,  and  the  latter  may  be 
used  without  serious  error.  (Tredgold's  Carpentry,  Arts.  57 
and  171.) 

The  pressures  in  an  equilibrated  frame  act  only  in  the 
axes  of  the  timbers  composing  the  frame,  and  these  carry 
the  effects  of  the  several  weights  on,  from  point  to  point, 
until  they  successively  arrive  at  A  and  B,  the  points  of 
final  support.  A  frame  thus  balanced  is  not,  however, 
stable,  for  if  subjected  to  additional  pressure,  however  small, 
at  any  one  of  the  points  of  support,  it  is  liable  to  derange- 
ment;  and  if  so  deranged  it  has  no  inherent  tendency  to 
recover  itself,  but  the  distortion  will  increase  until  total 
failure  ensues.  A  frame  thus  conditioned  is  therefore  said 
to  be  in  a  state  of  unstable  equilibrium ;  while  a  frame  of 
suspended  pieces,  as  in  Fig.  82,  is  said  to  be  in  a  state  of 
stable  equilibrium,  since,  if  disturbed  by  temporary  pressures, 
it  will  recover  its  original  position  when  they  are  removed. 

617. — Trussing-  a  Frame. — The  tendency  to  derangement 
and  consequent  failure,  .in  a  frame  such  as  Fig.  83,  can  be 
counteracted  by  additional  pieces  termed  braces,  located  in 
any  manner  so  as  to  divide  the  inclosed  spaces  into  triangles. 
For  example:  it  may  be  divided  into  the  triangles  ACS, 
CSD,  DST,  DTE,  RTF,  TFU  and  UFB.  If  these  additional 
pieces  be  adequate  to  resist  such  pressures  as  they  may  be 
subjected  to,  and  be  firmly  connected  at  the  joints,  the  frame 
will  thereby  be  rendered  completely  stable.  Treated  in 
this  way  the  frame  becomes  a  truss,  from  the  fact  that  it  has 
been  trussed  or  braced. 

618. — Forces  in  a  Truss  Graphically  Measured. — When 
a  frame  is  divided  into  triangles,  as  proposed  in  the  last 
article,  sometimes  three  or  more  pieces  meet  at  the  same 


418  FRAMED   GIRDERS.  CHAP.   XXII. 

point.  In  such  a  case,  owing  to  the  complexity  of  the 
forces,  it  becomes  difficult  to  trace,  and,  by  the  parallelogram 
of  forces,  to  measure  them  all.  Professor  Rankine,  in  his 
"  Applied  Mechanics,"  somewhat  extended  the  use  of  the 
triangle  of  forces  in  its  application  to  such  cases.  It  was 
afterward  more  fully  developed  and  generalized  by  Professor 
J.  Clerk  Maxwell  in  a  paper  read  before  the  British  Asso- 
ciation in  1867,  and  by  him  termed  ''Reciprocal  Figures, 
Frames  and  Diagrams  of  Forces  ;"  and  Mr.  R.  H.  Bow, 
C.E.,  F.R.S.E.,  in  his  "  Economics  of  Construction,"  has 
simplified  the  method  in  its  use,  by  a  system  of  reciprocal 
lettering  of  lines  and  angles.  By  Professor  Maxwell's 
method,  the  forces  in  any  number  of  pieces  converging  at 
one  point  are  readily  determined.  The  principle  involved 
is  very  simple,  and  is  this :  Construct  a  closed  polygon,  with 
lines  parallel  to  the  direction  of,  and  equal  in  length  to,  the 
amount  of  the  forces  which  in  the  framed  truss  meet  at  any 
point.  A  system  of  such  polygons,  one  for  each  point  of 
meeting  of  the  forces,  so  constructed  that  in  it  no  line 
representing  any  one  force  shall  be  repeated,  is  termed  a 
diagram  of  forces. 

619. — Example. —Let  Fig.  85  represent  a  point  of  con- 
vergence of  parts  of  a  framed  truss,  and  Fig.  86  be  its 
corresponding  diagram  of  forces,  in  which  latter  the  lines 
are  drawn  parallel  to  the  lines  in  Fig.  85.  Designate  a  line 
in  the  diagram  in  the  usual  manner,  by  two  letters,  one  at 
each  end  of  the  line,  and  indicate  the  corresponding  line  in 
Fig.  85  by  placing  the  same  two  letters  one  on  each  side  of 
the  line.  For  instance,  the  line  AB  of  86  is  parallel  with  that 
line  of  85  which  lies  between  the  spaces  A  and  B ;  and  so  of 
each  of  the  other  corresponding  lines.  In  Fig.  86  the  lines  are 
in  proportion  to  each  other,  respectively,  as  the  forces  in  the 
corresponding  lines  of  Fig.  85.  Thus  if  AD  (86),  by  any  scale, 


RECIPROCAL   FIGURES. 


419 


represents  the  force  in  the  line  between  A  and  D  (85),  then 
will  the  line  AB  equal  the  force  in  the  line  between  A  and 
B;  and  in  like  manner  for  the  other  lines  and  strains. 


620. — Another  Example. — Let  Fig.  87  represent  the  axial 
lines  of  the  timbers  of  a  roof  truss,  and  its  two  sustaining 
piers,  and  let  Fig.  88  be  its  corresponding  diagram  of  forces. 


FIG.  85. 


The  truss  being  loaded  uniformly,  the  three  arrows  (87) — 
one  at  the  ridge  and  one  at  the  apex  of  each  brace — repre- 
sent equal  loads.  Let  these  three  loads  be  laid  down  to  a 
suitable  scale  on  the  line  FJ  (88),  one  extending  from  F 
to  G,  another  from  *  G  to  ff,  and  the  third  from  H  to 
y.  The  half  of  these,  or  FE,  is  the  load  sustained  on  one  of 
the  supports  of  Fig.  87,  and  the  other  half,  EJ,  is  the  load 
upon  the  other  support.  In  Fig.  87  a  letter  is  placed  in  each 
triangle,  and  one  in  each  partly-enclosed  space  outside  of 
the  truss.  Each  line  of  the  figure  is  to  be  designated  by 
the  two  letters  which  it  separates;  thus,  the  line  between 
A  and  E  is  called  line  AE,  the  line  between  A  and  B 
is  called  line  AB,  etc.  In  Fig.  88  the  corresponding  lines 
are  designated  by  the  same  letters  ;  the  letters  here  being, 
as  usual,  at  the  ends  of  the  lines.  Also,  it  will  be  observed 
that  while  in  the  diagram  any  point  is  designated  in  the 
usual  way  by  the  letter  standing  at  it,  it  is  the  practice  in 


420 


FRAMED   GIRDERS. 


CHAP.   XXII. 


the  frame  to  designate  a  point  by  the  several  letters  which 
cluster  around  it ;  for  example,  the  point  of  support  FAE 
(Fig.  87).  The  diagram,  Fig.  88,  is  constructed  by  drawing  a 
line  parallel  with  each  of  the  lines  in  Fig.  87.  Thus  the  three 


FIG.  87. 


FIG.  88. 


forces  converging  in  Fig.  87  at  FAE,  the  left-hand  point  of 
support,  are  EF,  FA  and  AE\  and  in  Fig.  88  the  lines  EF, 
FA  and  AE,  drawn  parallel  with  the  corresponding  lines 
of  Fig.  87,  form  the  triangle  EFA.  Taking  the  forces  at 


DIAGRAM   OF  FORCES.  421 

point  GBAF,  87,  we  find  them  to  be  AF,  FG,  GB  and 
BA,  and  drawing,  in  88,  lines  parallel  with  these,  we  obtain 
the  quadrangle  AFGBA.  Again,  in  87  the  forces  at  point 
GBCH  are  BG,  GH,  HC  and  CB,  and  drawing,  in  88, 
lines  parallel  with  these,  we  obtain  the  closed  polygon 
BGHCB.  At  point  HCDJ  (87)  the  forces  are  CH,  HJ,  JD 
and  DC,  and,  in  88,  drawing  the  corresponding  lines  pro- 
duces the  quadrangle  CHJDC.  In  87  the  forces  at  point 
JED,  the  right-hand  support,  are  DJ,  JE  and  ED,  and 
in  88,  the  corresponding  lines  produce  the  triangle  DJE. 
In  87,  at  point  ABODE,  we  have  the  five  forces  EA,  AB, 
BC,  CD  and  DE,  and,  in  88,  the  corresponding  lines  give 
the  closed  polygon  EABCDE.  This  completes  the  dia- 
gram of  forces,  Fig.  88,  in  which  the  several  lines,  measured 
by  the  same  scale  with  which  the  three  loads  were  laid  off  on 
FJ,  are  equal  to  the  corresponding  forces  in  the  similar 
parts  of  Fig.  87. 

621. — Diagram  of  Force§. — In  this  manner  the  diagram 
of  forces  may  be  drawn  to  represent  the  strains  in  a  framed 
truss,  by  carefully  following  the  directions  of  Art.  618  ;  com- 
mencing by  first  laying  down  the  forces  which  are  known; 
from  which  the  ones  to  be  found  will  gradually  be  determined 
until  the  whole  are  ascertained. 

62 20 — Diagram    of   Forces— Order    of    Development. — 

When  more  than  two  of  the  forces  converging  at  any  one  point 
are  undefined  in  amount,  the  diagram  can  not  be  completed. 
Thus,  where  three  forces  converge  it  is  requisite  to  know  one 
of  them.  Of  four  forces,  two  must  be  known.  Of  five  forces, 
three  must  be  known. 

In  constructing  a  diagram,  the  first  thing  necessary  is  to 
establish  the  line  of  loads,  as  FJ  in  Fig.  88,  then  to  ascertain 
the  portion  of  the  total  load  which  bears  upon  each  of  the 


422  FRAMED    GIRDERS.  CHAP.    XXII. 

points  of  support,  AEF  and  JED  (Art.  56)  (one  half  on 
each  when  the  load  is  disposed  symmetrically),  and,  with  this, 
to  obtain  the  first  triangle  FEA.  From  this  proceed  up  the 
rafters,  or  to  where  the  points  of  convergence  have  the 
fewest  strains,  leaving  the  more  complex  points  to  be  treated 
later.  In  this  way  the  most  of  the  forces  which  affect  the 
crowded  points  will  be  developed  before  reaching  those 
points. 

623. — Reciprocal  Figures. — By  comparing  Figs.  87  and 
88,  we  see  that  the  lines  enclosing  any  one  of  the  lettered 
spaces  in  the  former  are,  in  the  latter,  found  to  radiate  from 
the  same  letter.  The  space  A  (87)  has  for  its  boundaries  the 
lines  AF,  AB  and  AE,  and  these  same  lines  in  88  radiate 
from  the  letter  A.  The  space  E  (87)  has  for  its  enclosing 
lines  EF,  EA,  ED  and  EJ,  and  these  same  lines  are 
found  to  radiate  from  the  point  E  (88).  Thus  the  diagram 
(88)  is  a  reciprocal  of  the  frame  (87). 

624. — Proportions  in  a  Framed  Girder. —  In  order  to 
treat  of  the  method  of  measuring  strains  in  trusses,  we  have 
digressed  from  the  main  subject.  Returning  now,  and  refer- 
ring to  the  relations  existing  between  a  girder  and  a  simple 
beam,  as  in  Arts.  603  to  605,  we  proceed  to  develop  the 
proportion  in  a  girder,  between  the  length  and  depth. 

A  girder,  as  generally  used,  serves  to  support  a  tier  of 
floor  beams  at  a  line  intermediate  between  the  walls  of  the 
building,  and  when  sustained  by  posts  at  points  not  over 
12  to  15  feet  apart,  may  be  made  of  timber  in  one  single 
piece.  But  when  a  girder  is  required  to  span  greater  dis- 
tances than  these,  it  becomes  requisite,  by  some  contrivance, 
to  increase  its  depth,  in  order  to  obtain  the  requisite 
strength.  An  increase  of  depth,  however,  may  interfere 
with  the  demand  tor  clear,  unobstructed  space  in  rooms  so 


PROPORTION   BETWEEN   DEPTH   AND   LENGTH.  423 

large  as  those  in  which  girders  are  required.  To  prevent 
this  interference,  the  depth  of  the  girder  should  be  the  least 
possible ;  although  diminishing  the  depth  will  increase  the 
cost ;  for  the  cost  will  be  in  proportion  to  the  amount  of 
material  in  the  girder,  and  this  will  be  in  proportion  to  the 
strains  in  its  several  parts,  and  the  strains  will  be  inversely 
as  the  height.  For  economy's  sake,  therefore,  as  well  as  for 
strength,  the  girder  should  have  a  fair  depth;  modified,  how- 
ever, by  the  demand  for  unobstructed  space. 

Where  other  considerations  do  not  interfere  to  prevent 
it,  the  depth  of  a  framed  girder  should  be  from  y1^  to  -J-  of 
the  length  ;  the  former  proportion  being  for  girders  25  feet 
long,  and  the  latter  for  those  125  feet  long.  If  these  two 
rates  be  taken  as  the  standard  rates,  respectively,  of  two 
girders  thus  differing  in  length  100  feet,  and  all  other 
girders  be  required  to  have  their  depths  proportioned  to 
their  lengths  in  harmony  with  these  standards,  their  rates 
will  be  regularly  graduated.  In  order  to  develop  a  rule 
lor  this,  let  the  two  standards  be  reduced  to  a  common 
denominator,  or  to  ^  and  /¥.  If  their  difference,  -£%,  be 
divided  into  100  parts,  each  part  will  equal 

i  i 

24  x  100       2400 

and  will  equal  the  difference  in  rate  for  every  foot  increase 
in  length  of  girder;  for  the  two  standards  are  100  feet 
apart.  The  scale  of  rates  thus  established  is  for  lengths  of 
girder  from  25  to  125  feet,  but  it  is  desirable  to  extend 
the  scale  back  over  the  25  feet  to  the  origin  of  lengths. 
To  do  this,  we  have  for  the  difference  in  rates  for  this  25 
feet,  25  x  ^Vo  =  *Hhr  =  ?V  Deducting  this  from  TL  (=  -&), 
the  rate  at  25  feet,  we  have  -fa  —  -fa  =  -fa,  the  rate  be- 
tween depth  and  length  at  the  origin  of  lengths  (if  such  a 
thing  were  there  possible).  Now  if  to  this  base  of  rates  we 


424  FRAMED    GIRDERS.  CHAP.    XXII. 


add  the  increase  (y^V^  of  the  length)  the  sum  will  be  the 
rate  at  any  given  length.  As  an  example:  What  should 
be  the  rate,  by  this  rule,  for  a  girder  125  feet  long  ?  For 
this  the  difference  in  rate  is  125  x  -^fa  =  ^WV  =  -fa.  Adding 
this  to  the  base  of  rates,  or  to  -fa,  as  above,  and  the  sum 
^  +  ^_  —  II  _  £}  the  required  rate.  This  is  one  of  the 
standard  rates.  The  other  standard  may  be  found  by  the 
rule  thus,  25  x  ^Vir  =  ^j-jfo  =  -fa.  Adding  this  to  the  base 
-A»  gives  -ft-  =  T^»  the  standard  rate.  We  have  therefore 
for  the  rate  at  any  length 

r=  .  -L  +  _!_/=  -!Zi_  1 


96    24OO      24OO    24OO     24OO 


r  = 

2400 


This  gives  the  rate  of  depth  to  length,  and  since  the  depth 
is  equal  to  the  rate  multiplied  by  the  length,  therefore 


2400 


d  =  (994.) 

24OO 

in  which  d  is  the  depth  between  the  axes  of  the  top  and 
bottom  chords,  and  /  is  the  length  (between  supports),  both 
being  in  feet. 

This  rule  will  give  the  proper  depth  of  a  girder,  and  may 
be  used  when  the  depth  is  not  fixed  arbitrarily  by  the  cir- 
cumstances of  the  case.  (See  Art.  572.) 

625.  —  Example.  —  What  should  be  the  depth  of  a  girder 
which  is    40    feet  long  between  supports? 
By  formula  (294.), 

+  40)x4x)  = 
2400 


ECONOMICAL   DEPTH.  425 

or  the  economical  depth  is    3    feet  and    7    inches,  measured 
from  the  middle  of  the  depths  of  the  top  and  bottom  chords. 
Again  :    What  should  be  the  depth  of  a  girder  which  is 
100    feet  long  in  the  clear  between  supports  ?     By  (294>\ 

(175  +  100)  x  IPO 
a  =  --  —  ----  —  1  1  -450^ 
2400 

or  the  depth  between  the  axes  of  the  chords  should  be    n 
feet     5f     inches. 

A  girder  125  feet  long  would  by  this  rule  be  15  feet 
7-J-  inches,  or  -J-  of  its  length,  in  depth  ;  while  a  girder 
25  feet  long  would  be  2  feet  and  i  inch  deep  between 
the  axes,  or  T^  of  the  span. 

626.  —  Trussing,  in  a  Framed  Girder.  —  One  object  to  be 
obtained  by  the  trussing  pieces  —  the  braces  and  rods  —  is  to 
transmit  the  load  from  the  girder  to  the  abutments.  The 
braces  and  rods  forming  the  trussing  may  be  arranged  in  a 
great  variety  of  ways  (see  Bow's  Economics  of  Construction), 
but  that  system  is  to  be  preferred  which  will  take  up  the 
load  of  the  girder  at  proper  intervals,  and  transmit  it  to  its 
two  supports  in  the  most  direct  and  economical  manner. 

Just  which  of  the  great  number  of  systems  proposed  will 
the  more  nearly  perform  these  requirements  it  will  perhaps 
be  somewhat  difficult  to  determine,  but  the  one  in  which  the 
struts  and  ties  are  arranged  in  a  chain  of  isosceles  triangles 
is  quite  simple,  and  offers  advantages  over  many  others.  It 
is  therefore  one  which  may  be  adopted  with  good  results. 


627.  —  Plaaniiis  a  Framed  Girder.  —  After  fixing  upon 
the  height  (Art.  624),  the  next  point  is  as  to  the  number  of 
panels  or  bays.  These  should  be  of  such  length  as  to  afford 
points  of  support  at  suitable  intervals  along  the  girder,  and 
the  rods  and  struts  should  be  placed  at  such  an  angle  as  will 


4^6  FRAMED    GIRDERS.  CHAP.   XXII. 

secure  a  minimum  for  the  strains  in  the  truss.  To  set  the 
braces  and  ties  always  at  the  same  angle,  would  result  in  fur- 
nishing points  of  support  at  intervals  too  short  in  the  girders 
of  short  span,  and  too  long  in  those  of  long  span.  So  also,  if 
the  width  of  a  bay  be  a  constant  quantity,  there  would  be 
too  great  a  difference  in  the  angles  at  which  the  rods  and 
struts  would  be  placed.  To  determine  the  number  of  bays, 
so  as  to  avoid  as  far  as  practicable  these  two  objections—  first, 
we  have  the  number  of  the  bays  directly  as  the  length  of  the 
truss  and  inversely  as  the  depth,  arid,  second  (to  vary  this  pro- 
portion as  above  suggested),  we  may  deduct  from  this  result 
a  quantity  inversely  proportioned  to  the  length  of  the  girder. 
Combining  these,  we  have,  n  being  the  number  of  bays, 

/        120—  / 

n  —  ~j  -- 
d  c 

and  by  substituting  for    d    its  value  as  in  formula 

I  120-1 


2400 

2400      1  20—  / 
"  175  +  '  ~    ~~c~ 

in  which     /    is  the  length  of  the  girder  in  feet,  and     c    is  a 
constant,  to  be  developed  by  an  application  to  a  given  case. 
To  this  end  we  have,  from  the  last  formula, 

1  20—  /_     2400 

—    =l7sT7- 

I2O  —  / 


2400 


With     n  =4-5     and     /=  20,  we  have 


DETERMINING   THE   NUMBER   OF   BAYS.  427 


120—20  100 

C  =  --  -  =  -  ;  -  =  12  •  SO7 

2400         _  12-308—4.5 

175  +  20  ~~ 


or,  say     c  =  12-8  ;     and  with  this  value 


I75  +  /        12-8 

which  is  a  rule  for  determining  the  number  of  bays  in  a  truss, 
when  not  determined  arbitrarily  by  the  circumstances  of  the 
case,  and  when  the  height  of  the  girder  is  obtained  as  in 
formula  (294-}.  In  the  resulting  value  of  ;/,  the  fraction 
over  a  whole  number  is  to  be  disregarded,  unless  greater 
than  £,  in  which  latter  case  unity  should  be  added  to  the 
whole  number. 


628.  —  Example.  —  What  should  be  the  number  of  bays  in 
a  truss    80    feet  long  ? 

Here     /  =  80,    and,  by  the  formula, 


2400         1  20—  80  _  2400        40 
n=  i75  +  8o~~T^~8~    :  "255"  ~~~i2T8  - 

or   the   required  number  of  bays  is  six  ;    disregarding   the 
decimal     •  287     because  it  is  less  than     £. 


629.  —  Example.—  How    many   bays    are    required   in   a 
girder     no    feet  long? 

Here    /=  no,    and,  by  formula 


2400          1  20—  no 
=  ~ 


or,  adding  unity  for  the  decimal     -64,    the  number  required 
is     8. 


FRAMED    GIRDERS.  CHAP.    XXII. 

630.  —  Number  of  Bays  in  a  Framed  Oirder.  —  According 
to  the  above  rule,  the  number  of  bays  or  panels  required  in 
framed  girders  of  different  lengths  is  as  follows  : 

Girders  from       20     to      59    feet  long  should  have    5    bays. 
«  "          59     «       85       "       "          "          "        6       " 

85     "     107      "      "          "         "       7       " 


"          "         85     "     107      "      "          "         " 

«  «        IQy     tt     I2y      «       u          «          u        8 

"  "        127      "     146       "       "  "          "        9 


In  cases  where  the    length   exceeds     120     feet,  the  quantity 

of  the  formula  to  be  deducted     f  -        -  j     becomes  a  nega- 

V    12-5  i 

tive  quantity,  and,  since  deducting  a  negative  quantity  is 
equivalent  to  adding  a  positive  one,  the  result  may  be  added, 
thus: 

120—144  —24 


631. — Forces  in  a  Framed  Oirder. — Let  Fig.  89  represent 
the  axial  lines  of  a  framed  girder,  or  the  imaginary  lines 
passing  through  the  axes  of  the  several  pieces  composing  the 
frame.  Let  the  load,  equally  distributed,  be  divided  into  six 
parts,  one  of  which  acts  at  the  apex  of  each  lower  triangle. 
We  may  notice  here  that  in  a  truss  with  an  even  number  of 
lower  triangles,  as  in  Fig.  89,  there  is  an  even  number  of 
loads,  one  half  of  which  are  carried  by  the  struts  and  rods  to 
one  point  of  support,  and  the  other  half  to  the  other  support. 
Thus  the  load  PQ,  at  point  PEFGQ,  is  sustained  by  the 
top  chord,  and  by  the  strut  EF.  The  portion  passing  down 
this  strut  is  carried  by  the  rod  DE  up  to  the  top  chord, 
and  thence,  together  with  the  load  OP,  at  point  OCDEP, 
down  by  the  strut  CD  to  the  bottom  chord.  This  accu- 
mulated load  is  carried  by  the  rod  BC  up  to  the  top  chord, 


DETERMINING  THE    PRESSURES.  429 

and  thence,  with  the  addition  of  the  last  load  AO,  at 
ABCO,  finally  reaches,  through  the  strut  AB,  the  point 
of  support  for  that  end  of  the  truss.  The  three  weights  on 
the  other  side  are  in  like  manner  conveyed  to  MNT,  the 
other  point  of  support.  We  here  see  the  manner  in  which, 
in  a  framed  girder  upon  which  the  load  is  uniformly  dis- 
tributed, one  half  is  carried  by  the  trussing  pieces  to  each 
point  of  support. 

632. — Diagram  for  the  above  Framed  Girder. — Fig.  90  is 

a  diagram  constructed  as  per  Arts.  619  and  620,  and  repre- 

sgpfo-.iK^v^ ,  •  trie  sides  ot  which  measure  the  torc&>  5».  89. 

s  converging  at  the  point    BCDT    of  89.     The  next  ir 

er  is  the  point    OCDEP.      Of  the  five  forces  concentrat 

here,  we  already  have,  in  Fig.  90,  three,    PO,     OC    an 

'.      To  find  the  other  two,  draw  from    D    the  line    D 

-allel  with   the   line    DE    of  89,   and  from    P    draw    PL 

rallel  with  line    PE    of  89.     These  two  lines  will  meet  a 

and    complete  the  polygon    POCDEP,    which  measures 

3    forces    in    the    lines    concentrating   at    point     OCDEP 

oceeding  now  to  the  point    DEFT  of  Fig.  89,  we  find  foui 

•ces,    two   of  which,      TD    and    DE.,    are    already    deter 

ned.     For   the    other   two,    draw    from     E    the    line    El 

rallel  with     EF    in  89,  and  from    T,     TF    parallel  with  th 

e     TF  of  89.     These  two  lines  meet  in    F   and  complete 

3    polygon     TDEFT,    which  measures   the   forces  in   the 

es  converging  at  the  point    DEFT    of  Fig.  89.     The  nex 

order  is  the    >oint    PEFGQ    in  89,  where  five   IWe 

FIG.  90. 

To  construct  this  diagram,  we  proceed  as  follows:  Upon  the 
vertical  line  AN  lay  off  the  several  distances  AO,  OP, 
PQ,  QR>  RS  and  SN\  each  equal  by  any  convenient 


430  FRAMED    GIRDERS.  CHAP.    XXII. 

scale  to  one  of  the  six  equal  loads  resting  upon  the  top  of  89. 
The  load  at  the  apex  of  the  triangle  ./?,  or  point  ABCO 
(89),  is  placed  from  A  to  O  in  90.  The  load  OP,  at  point 
OCDEP  (89),  is  placed  from  O  to  P  in  90 ;  and  so  on  with 
the  other  loads.  The  other  lines  of  Fig.  90  are  obtained  by 
drawing  them  parallel  with  the  corresponding  lines  of  Fig. 
89,  as  per  directions  in  Art.  618.  Commencing  at  the  point 
ABT  (89),  we  draw  (in  90)  three  lines  parallel  with  the  direc- 
tion of  the  forces  at  that  point.  The  first  of  these  is  the  ver- 
tical pressure  upon  the  point  of  support  ABT,  which  in 
this  case  equals  one  half  the  total  load,  or  AQ,  or  AT  of 

eu  ..    uvj     UC    UCLlUCLCci         i  Q       /          ~   "  ~ 

re  quantity,  and,   since  deducting  a  negative  quantity 
[uivalent  to  adding  a  positive  one,  the  result  may  be  adde 
us: 

• 
120— 


631. — Force§  in  a  Framed  Girder. — Let  Fig.  89  represen 
^  axial  lines  of  a  framed  girder,  or  the  imaginary  line 
ssing  through  the  axes  of  the  several  pieces  composing  th 
ime.  Let  the  load,  equally  distributed,  be  divided  into  si: 
rts,  one  of  which  acts  at  the  apex  of  each  lower  triangle 
e  may  notice  here  that  in  a  truss  with  an  even  number  o 
wer  triangles,  as  in  Fig.  89,  there  is  an  even  number  c 
ids,  one  half  of  which  are  carried  by  the  struts  and  rods  t 
e  point  of  support,  and  the  other  half  to  the  other  suppon 

FIG.  90. 

Fig.  90.  Next,  from  T,  draw  the  horizontal  line  Tfiy  and 
from  A,  the  inclined  line  AB,  parallel  with  the  brace 
AB  of  89.  These  two  lines  meet  at  B,  and  we  have  the  tri- 
angle ABT,  representing  the  three  forces  converging  at. 
the  point  of  support  ABT.  For  the  four  forces  at  the  point 


CONSTRUCTING   DIAGRAM    OF    FORCES.  43! 

ABCO  in  89,  we  proceed  as  follows  :  We  already  have  the 
forces  AO  and  AB.  From  B  in  90,  draw  BC  parallel 
with  the  rod  BC  of  89 ;  and  from  O  in  90,  draw  OC  par- 
allel with  OC  of  89.  These  two  lines  intersect  at  C,  com- 
pleting the  polygon  ABCOA,  the  sides  of  which  are  in  pro- 
portion as  the  forces  in  the  several  lines  converging  at  the 
point  ABCO  of  Fig.  89.  Proceeding  to  the  point  BCDT 
(Fig.  89),  we  find,  of  the  four  forces  converging  there,  that  two 
are  already  drawn,  777  and  BC.  From  C  draw  CD 
parallel  with  the  brace  CD  of  Fig.  89  ;  and  from  T  draw 
TD.  These  two  lines  will  meet  at  D  and  complete  the  poly- 
gon TBCDT,  the  sides  of  which  measure  the  forces  in  the 
lines  converging  at  the  point  BCDT  of  89.  The  next  in 
order  is  the  point  OCDEP.  Of  the  five  forces  concentrat- 
ing here,  we  already  have,  in  Fig.  90,  three,  PO,  OC  and 
CD.  To  find  the  other  two,  draw  from  D  the  line  DE 
parallel  with  the  line  DE  of  89,  and  from  P  draw  PE 
parallel  with  line  PE  of  89.  These  two  lines  will  meet  at 
E  and  complete  the  polygon  POCDEP,  which  measures 
the  forces  in  the  lines  concentrating  at  point  OCDEP. 
Proceeding  now  to  the  point  DEFT-  of  Fig.  89,  we  find  four 
forces,  two  of  which,  TD  and  DE.,  are  already  deter- 
mined. For  the  other  two,  draw  from  E  the  line  EF 
parallel  with  EF  in  89,  and  from  T,  TF  parallel  with  the 
line  TF  of  89.  These  two  lines  meet  in  F  and  complete 
the  polygon  TDEFT,  which  measures  the  forces  in  the 
lines  converging  at  the  point  DEFT  of  Fig.  89.  The  next 
in  order  is  the  point  PEFGQ  in  89,  where  five  lines  con- 
verge. The  forces  in  three  of  these  we  have  already — namely, 
QPy  PE  and  EF.  Draw  from  F  a.  line  parallel  with  the 
line  FG  of  89,  and  from  Q  a  line  parallel  with  QG  -of 
89.  These  two  intersect  at  G  and  complete  the  polygon 
QPEFGQ,  which  gives  the  forces  in  the  lines  around  the 
point  PEFGQ  of  Fig.  89. 


432  FRAMED    GIRDERS.  CHAP.   XXII. 

In  this  last  proceeding  we  meet  with  a  peculiarity.  The 
line  FG  has  no  length  in  Fig.  90.  It  commences  and  ends 
at  the  same  point,  since  G  is  identical  with  F.  This 
seems  to  be  an  error,  but  it  is  not.  It  is  correct,  for  an  ex- 
amination of  Fig.  89  will  show  that  the  two  inclined  lines 
meeting  at  the  foot  of  the  triangle  G  do  not  assist  in  carry- 
ing the  weights  upon  the  top  chord,  and  may  therefore,  in  so 
far  as  those  weights  are  concerned,  be  dispensed  with,  so 
that  the  space  occupied  by  the  three  triangles  F,  G  and 
H  may  be  left  free,  and  be  designated  by  one  letter  only  in- 
stead of  three.  In  place  of  five,  there  are  in  fact  only  four 
forces  meeting  at  the  point  PEFGQ,  and  these  four  are  rep- 
resented in  Fig.  90  by  the  polygon  QPEFQ. 

The  above  analysis  is  in  theory  strictly  correct,  and  yet 
in  practice  it  is  not  so,  for  in  such  cases  as  this  there  is  al- 
ways more  or  less  weight  on  the  lower  chord  at  the  middle 
point.  If  nothing  more,  there  is  the  weight  of  the  timber 
chord  itself,  and  this  should  be  considered. 

In  Art.  634  a  truss  with  weights  at  the  points  of  each 
chord  will  be  found  discussed,  and  the  facts  as  found  in  prac- 
tice there  developed. 

The  construction  of  one  half  of  the  diagram  (Fig.  90)  has 
now  been  completed.  The  other  half  is  but  a  repetition  of 
it  in  reversed  order,  and  need  not  here  be  shown  in  detail. 
In  drawing  the  lines  for  the  latter  half,  it  will  be  seen  that  the 
point  H  is  identical  with  the  point  F,  and  that  K  and 
M  coincide  respectively  with  D  and  B. 


633. — Gradation  of  Strains  in  Chords  and  Diagonals. — 

In  considering  the  forces  shown  in  Ftg.  90,  we  find  that  those 
in  the  chords  increase  towards  the  middle  of  the  girder,  while 
the  forces  in  the  diagonals  decrease  towards  the  middle. 
Thus,  in  Fig.  90,  of.  the  lines  representing  the  upper  chord, 


LOADS  ON  EACH  CHORD.  433 

PE  is  longer  than  OC,  and  QG  is  longer  than  PE,  in- 
dicating a  corresponding  increase  in  the  lines  OC,  PE  and 
QG  of  89.  So  in  the  lower  chord,  we  have  a  successive  in- 
crease of  forces,  as  seen  in  a  comparison  of  the  lengths  of  the 
lines  TB,  TD  and  TF  of  Fig.  90,  representing  the  chord 
at  the  several  bays  B,  D  and  F  of  Fig.  89.  The  diagonal 
lines  AB,  CD  and  EF  in  90  show  decreasing  forces  in 
the  diagonals  AB,  CD  and  EF  in  89 — decreasing  towards 
the  middJa/to  a  point  beyond  the  middle  of  tnYi/P  remem- 
ber wVphe  remainder  of  Fig.  92  may  be  traced  for  thevledge 
of  thjf  9I^  by  a  continuance  of  the  process  used  in  tracing6  di- 
ag"olhalf.  Since,  in  this  instance,  the  loading  and  plan  of  the 
dialer  are  symmetrical,  and  hence  the  several  forces  in 
2S  of  one  half  of  the  girder  respectively  equal  to  those 
3  other  half,  the  lines  of  the  diagram  as  laid  down  for 
e  may  be  used  for  the  other  half.  Let 

Fi±  rre- 

sp<  -am, 

635. — Gradation  of  Strains  in  Chord  §  and  Diagonal    . 

we  the 

e  gradation  of  the  forces  in  Fig.  92  may  (as  was  remai 

.    Art.  633)  be  observed  in  the  diagonals  representing 
.28   KA,   AB,   BC,    CD    and    DE,  which  diagonals  decrt 

,    m  the  end  towards  the  middle  of  the  girder  ;  and  also  \. 
.  ^  lines  representing  the  chords   A  U,   BL,    CT,    DM  an( 

which  gradually   increase   from   the    end    towards    the 

pai  J 

idle. 

two  J. 

In  tl  ^~d  being  symmetrically  ui^^      ,     1-  cwo 

parts  are  equal,  or  KU  =  UV.  From  U  and  K  draw  lines 
parallel  to  the  corresponding  lines  UA  and  KA  (91).  These 
will  meet  at  A  and  complete  the  triangle  of  forces  for  the 
point  A  UK  of  Fig.  91.  From  A  in  92  draw  the  Une  AB, 
and  from  L  the  line  LB.  These  meet  at  B  and  com- 
plete the  polygon  KLBAK  for  the  forces  at  the  point 
KABL  of  91.  Starting  from  U,  set  off  upon  the  vertical 


434  FRAMED    GIRDERS.  CHAP.    XXIT. 

line  KV  the  several  distances  UT,  TS,  SR  and  RQ, 
respectively  equal  to  the  several  loads  UT,  TS,  SR  and 
RQ  as  found  in  91.  For  the  forces  at  the  point  ABCTU, 
draw  the  line  BC  from  B,  and  the  line  TC  from  T, 
each  parallel  with  its  corresponding  line  in  91.  These  lines 
meet  at  C  and  complete  the  polygon  ABCTUA,  which 
gives  the  forces  converging  in  the  point  UABCT. 


occupied  by  the  three  triangles    .  , 
y  be  left  free,  and  be  designated  by  one  letter  OL 
•  of  three.     In  place  of  five,  there  are  in   fact  only 
3S  meeting  at  the  point  PEFGQ,    and  these  four  are  i 
nted  in  Fig.  90  by  the  polygon     QPEFQ. 
The  above  analysis  is  in  theory  strictly  correct,  and  } 
>ractice  it  is  not  so,  for  in  such  cases  as  this  there  is  i 
s  more  or  less  weight  on  the  lower  chord  at  the  midd 
t.     If  nothing  more,  there  is   the  weight  of  the  timb' 
id  itself,  and  this  should  be  considered. 
i  Art.  634  a  truss  with  weights  at  the  points  of  eac 
3  will  be  found  discussed,  and  the  facts  as  found  in  pra 
ythere  developed. 

/The  construction  of  one  half  of  the  diagram  (Fig.  90)  h 
jw  been  completed.     The  other  half  is  but  a  repetition 
/  in  reversed  order,  and  need  not  here  be  shown  in  deta 
(n  drawing  the  lines  for  the  latter  half,  it  will  be  seen  that  t 
oint    H    is  identical  with  the  point    Fy    and  that    K   a 
Coincide  re«—  -  FIG.  92, 

For  the  point  LBCDM,  draw  from  C  the  line  CD, 
and  from  M  the  line  MD^  each  parallel  with  its  corre- 
sponding line  in  91.  These  lines,  meeting  in  D,  complete 
the  polygon  MLBCDM^  which  gives  the  forces  surround- 
ing the  point  LBCDM.  For  the  point  TCDES,  draw  from 
D  the  line  D£,  and  from  5  the  line  SE,  respectively 


GRADATIONS   OF   STRAINS.  435 

parallel  with  the  corresponding  lines  of  Fig.  91.  They  will 
meet  at  E  and  complete  the  polygon  TCDEST,  which 
measures  the  forces  around  the  point  TCDES.  For  the 
point  MDEFNy  draw  from  E  the  line  EF,  and  from  N 
the  line  NF,  parallel  with  EF  and  NF  of  91 ;  and  they, 
meeting  at  F,  will  complete  the  polygon  MDEFNM,  thus 
giving  the  forces  converging  at  the  point  MDEFN. 

The  correspondence  of  lines  in  the  two  figures  has  now 
been  traced  to  a  point  beyond  the  middle  of  the  framed  gir- 
der. The  remainder  of  Fig.  92  may  be  traced  for  the  other 
half  of  91,  by  a  continuance  of  the  process  used  in  tracing  the 
first  half.  Since,  in  this  instance,  the  loading  and  plan  of  the 
girder  are  symmetrical,  and  hence  the  several  forces  in  the 
lines  of  one  half  of  the  girder  respectively  equal  to  those  in 
the  other  half,  the  lines  of  the  diagram  as  laid  down  for  the 
one  may  be  used  for  the  other  half. 

635a — Gradation  of  Strains  in  Chords  and  Diagonals. — 

The  gradation  of  the  forces  in  Fig.  92  may  (as  was  remarked 
in  Art.  633)  be  observed  in  the  diagonals  representing  the 
lines  KA,  AB,  BC,  CD  and  DE,  which  diagonals  decrease 
from  the  end  towards  the  middle  of  the  girder ;  and  also  in 
the  lines  representing  the  chords  A  U,  BL,  CT,  DM  and 
ES,  which  gradually  increase  from  the  end  towards  the 
middle. 


636. — Strains  Measured  Arithmetically. — Let  Fig.  93 
represent  a  framed  girder,  in  which  the  loads  are  symmetri- 
cally placed,  and  where  L  is  put  for  the  load  on  each  point 
of  bearing  of  the  upper  chord,  and  N  for  that  suspended 
at  each  bearing  point  of  the  lower  chord.  Let  a  represent 
the  vertical  height  of  the  girder,  c  the  length  of  a  diagonal, 
and  b  the  base  of  the  triangle  formed  with  c  and  a. 


436 


FRAMED    GIRDERS. 


CHAP.    XXII. 


637. — Strains  in  the  Diagonals. — To  analyze  these,  we 
commence  at  the  middle  of  the  girder.  There  being  an  odd 
number  of  loads  upon  the  upper  chord,  one  half  of  the 


Fin.  93- 

central  one,  Z,  is  carried  at  Q,  one  of  the  points  of  sup- 
port, and  the  other  half  at  F,  the  other  point.  The  effect 
of  this  upon  .the  brace  MC  may  be  had  from  the  relation  of 
the  sides  of  the  triangle  abc,  for 


a  :  c  : :  \L  :   —L 
2a 


2a  :  c  : :  L  :   —L 
2a 


D 


equals  the  strain  in  the  diagonal ;  or,  when    W  equals  the 
vertical  load,  equals    JZ, 


D  =  W- 
a 


(296.) 


The  vertical  effect  of  this  at  M  is  JZ-,  the  same  as  it  is 
at  C.  This  amount,  added  to  the  suspended  load  N  at  M, 
equals  \L  +  A7",  equals  the  total  vertical  force  acting  at  M. 
This  is  sustained  by  the  lines  MK  and  BM,  the  latter 
standing  at  the  same  angle  with  MK  as  did  MC.  Hence 
the  effect  upon  the  diagonal  is 


STRAINS   IN   THE   DIAGONALS.  437 

equals  the  strain  on  the  diagonal  BM ;  and  the  vertical 
effect  at  M  is  equal  to  \L-\-  N.  Adding  this  to  the  load  on 
the  top  chord  at  B,  the  sum,  fZ+TV,  is  the  total  load  at 
B,  and  it  is  supported  by  the  forces  in  the  lines  PB  and 
BC,  constituting,  with  the  weight,  three  forces,  acting  in 
the  directions  of  the  three  sides  of  the  triangle  abc.  The 
effect  in  the  diagonal  BP  is  therefore,  as  before,  the  load 


into  the  ratio     — ,    or    (IL  +  N)— .     The  vertical  effect  of  this 

a  V2  }  a 

at  P  is  equal  to  the  vertical  effect  at  B,  or  %L  +  N.  Add- 
ing to  it  the  load  N  at  P,  their  sum,  J-Z  4-  2N,  is  the  total 
vertical  effect  at  P ;  and,  as  before,  the  effect  of  this  on  the 

diagonal  AP  which  carries  it  is  (fZ  +  2N)—,    with  a  vertical 

effect  at  A  of  f  Z  4-  2N,  the  same  as  at  P.  Adding  the 
load  Z,  at  A,  the  sum,  §L+2N,  equals  the  total  vertical 
pressure  at  A.  This  is  sustained  by  the  forces  in  the  lines 
QA  and  AB,  which,  with  the  weight,  act  in  the  direction 
of  the  sides  of  the  triangle  abc,  and  therefore  the  effect  in 

the  diagonal,   as  before,  is    (§-£4-2^)—,    while   the  vertical 

a 

effect  of  this  at     Q    is  equal  to  the  same  effect  as  at     A,    or 


Thus,  the  loads  on  half  the  girder  have,  one  by  one,  been 
picked  up  and  brought  along,  step  by  step,  until  they  are 
finally  received  upon  Q,  their  point  of  support  at  one  end 
of  the  girder. 

It  will  be  observed  that  this  accumulated  load,  %L+2N, 
coincides  with  the  sum  of  the  loads  as  seen  upon  one  half  of 
the  figure,  that  is,  to  the  2-J-  loads  on  the  top  chord  and  the 
two  loads  suspended  from  the  bottom  chord. 

638. — Example. — Let  it  be  required  to  show  the  strains 
in  the  diagonals  of  a  framed  girder  50  feet  long,  of  five 
bays  and  4^  feet  high. 


438 


FRAMED    GIRDERS. 


CHAP.   XXII. 


Here  b,  the  base  of  the  measuring  triangle,  is  equal  to 
|~o  =  5,  and  a,  its  height,  equals  the  height  of  the  girder, 
equals  4-5  ;  and  <:,  the  hypothenuse  of  the  right-angled 
triangle,  is  therefore 


c— 


—  1/20-25  +  25  =6-7268 


The  load  L  upon  each  point  of  the  upper  chord  is  10,000 
pounds,  while  N,  the  suspended  load  at  each  point  of  the 
bottom  chord,  is  2500  pounds. 


FIG.  93. 
The  strain  upon  the  diagonals  is,  by  formula  (296. \ 


The  load  on    CM  is   %L,   and  therefore  the  strain  in  the 
diagonal    CM  is 


c                       6-7268 
D,  =  L~  =  loooo  x  — r =  74741  pounds. 


The  strain  in  the  diagonal   MB   is 

c                              ,6-7268 
Da  —  ($L  +  M)  -  =  (5000  +  2500)  - =  1 121  it  pounds. 


STRAINS   IN   LOWER  CHORD.  43C 

The  strain  in  the  diagonal   BP  is 

c  ,6-7268 

D3  —  (f-Z,  +  N )  -  =  (i  5000  +  2500)  -         -  =  261 59!  pounds. 
'a  4-5 

The  strain  in  the  diagonal   PA    is 

D&  =  (4Z,  +  2N)  -  =  (i  5000  +  5000)  — —  -  =  29896!  pounds ; 
a  4*5 

and  the  strain  in  the  diagonal   A  Q   is 
D5  =  (%L  +  2N)  -  —  (25000+  5000)  -  -— —  =  448451  pounds. 


639.  —  Strains  in  tlie  L,ower  Chord.  —  From  the  measuring 
triangle  abc  of  Fig.  93  we  have 

a  :  b  ::    W  :  H 

H=W-  (MT-) 

a 

in  which  H  is  the  strain  in  the  horizontal  lines  due  to  W 
the  weight  ;  and  with  this  formula  we  may  ascertain  the 
horizontal  forces  in  the  chords  of  the  girder. 

First.  In  the  lower  chord.  At  the  point  Q  we  have,  for 
W  in  the  formula,  one  half  the  total  load,  or  (JZ,  +  2N\ 
and  therefore 


equals  the  horizontal  strain  in    QP. 

For  the  next  bay,  PM,  we  have,  for  W,  the  same  amount, 
plus  that  caused  by  the  thrust  in  the  strut  BP,  plus  that 
due  to  the  tension  in  the  rod  AP.  These  three  amounts 


440 


FRAMED   GIRDERS. 


CHAP.   XXII. 


are    respectively     f  Z,  +  2N,     f  £  +  N     and      f  L  +  2N,      and 
their  sum  is 

(f£  -f  27V)  +  (f  £  4-  JV)  f  (f  £  -f  2^0  =  JT  =  -y-£  +  5^ 

equals  the  total  weight  causing-  horizontal  strain  in   PM. 
From  this,  the  horizontal  strain  in    PM   is 


.. 

For  the  third,  or  middle  bay,    MK,    we  have  the  weight 
the  same  as  for    PM,    together    with   that  coming  from  the 


1    . 


FIG.  93. 

thrust  of  the  strut  CM,  and  from  the  tension  of  the  rod  BM. 
These  three  weights  are  ty-L  -f  $N>  \L  and  %L+  N,  or 
together, 

(V£  +  $N)  +  \L  +  (f£  +  N)~  W=±£L  +  6A? 
and  for  the  horizontal  strain  in   MK  we  have 


This  completes  the  strains  in  the  lower  chord,  for  those  of 
the  other  end  are  the  same  as  these. 


640. — Strains  in  the  Upper  Chord. — For  the  first  bay, 
AB,  there  are  two  compressions,  namely:  that  due  to  the 
reaction  from  the  strut  AQ,  and  that  from  the  tension  in 


STRAINS   IN   UPPER   CHORD.  441 

the  rod  AP.  The  weight  causing  thrust  in  the  strut  is 
equal  to  half  the  total  load,  or  fZ  4-  27V,  and  the  weight 
causing  tension  in  the  rod  is  f  Z  +  2N ;  or,  together,  we 
have  for  the  weight  4Z  4-  47V;  and  for  the  compression  in 
AB, 

H'  =  (AL  4-  47V)  — 
d 

For  the  second  bay,  BC,  we  have  this  same  thrust,  plus 
that  due  to  the  reaction  of  the  strut  PB,  plus  that  due  to 
the  tension  in  the  rod  BM.  The  three  weights  are  4Z  +  4/V, 
fZ  +  N  and  -JZ  4-  N,  and  their  sum  is 

(4Z  4-  47V)  4  (f  Z  4-  TV)  4-  (JZ  4-  TV)  =  6Z  4-  67V 

and  the  horizontal  compression  in   BC  is 

£ 

77"  =  (6Z  +  6yV)- 
i? 

641. — Example. —  What  are  the  horizontal  strains  in  a 
girder  of  five  bays,  it  being  50  feet  long  and  4^  feet  high, 
and  having  10,000  pounds  resting  upon  each  bearing  point 
of  the  upper  chord,  and  2500  pounds  at  each  point  of  sus- 
pension in  the  lower  chord? 

Here,    in   the  measuring   triangle    abc,    b  —  -f-^-  =  5    and 

a  =  4-  5  ;    from  which    -  =  -—  =  i4.      Hence,  for  each  hor- 

a       4-5 

izontal  strain,  we  have 


•    Now,  in  the  lower  chord,  we  have,  as  in  Art.  639,  for  the 
bay    QP,    the  weight 

W=    L-\-2N  and,  therefore, 


Hf  =  -1/  [(2^-  x  10000)  +  (2  x  2500)]  =  33333-}    pounds  ; 
equals  the  horizontal  tension  in     QP. 


442  FRAMED    GIRDERS.  CHAP.    XXII. 

For  the  next  bay,    PM,    we  have  for  the  weight,  as  per 
Art.  639, 

W  =  V1  7-  f  5^  and,  therefore, 

-ft  =  -V-KSi  x  10000)  +  (5  x  2500)]  =  75000    pounds  ; 

equals  the  horizontal  tension  in    PM. 

For  the  third,   or  middle   bay,    MK,    for  the  weight,  as 
per  Art.  639,  we  have 


W=  ifL  -f  6N  and,  therefore, 

Hs  =  -V-[(6i  x  i  oooo)  -h  (6  x  2500)]  =  88888|    pounds  ; 

equals  the  horizontal  tension  in    MK. 

This  completes  the  work  for  the  lower  chord,  as  the  ten- 
sions in  the  other  half  are  the  same  as  those  here  found  for 
this. 

In  the  upper  chord  the  weight  causing  compression  in  the 
first  bay,  A£,  is,  as  per  the  last  article, 

W  —  4L  i  4N  and,  therefore, 

H'  =  -¥[(4><  10000)  +  (4x2500)]  =  555551    pounds; 

equals  the  horizontal  compression  in    AB. 

For  the  next  bay,  BC,  for  the  weight  causing  compres- 
sion we  have,  as  per  last  article, 

W  —  6L  -f  6N  and,  therefore, 

H"  =  W6  x  10000)  -f-  (6  x  2500)]  =  83333^    pounds  ; 

equals  the  horizontal  compression  in    BC. 


HORIZONTAL   STRAINS — TENSION.  443 

This  completes  the  strains  for  the  upper  chord.     Tabu- 
lated, these  several  horizontal  strains  stand  thus  : 
For  the  lower  chord : 

In    QP   and    JF    the  strains  are    33,333^    pounds. 

"    PM      "      KJ     "         "         tk       75,ooo 

"    MK  "    strain  is         88,888|         " 

For  the  upper  chord  : 

In    AB    and    DE    the  strains  are    55,555|    pounds. 
"     BC       "      CD      "         "         "      83,333^ 

To  test  the  correspondence  of  these  results  with  those 
shown  by  the  graphic  method  in  Figs.  91  and  92,  the  student 
may  make  diagrams  with  the  given  figures  at  a  scale  as  large 
as  convenient,  giving  to  L  and  N  the  proportions  above 
assigned  them,  namely,  L  —  4.N,  and  making  the  bays  with 
a  base  of  10  and  a  height  equal  to  4-5.  The  results  ob- 
tained should  approximate  those  above  given,  in  proportion 
to  the  accuracy  with  which  the  diagrams  are  made. 


642. — Resistance  to  Tension. — Only  in  so  far  as  tension 
is  incidental  to  the  transverse  strain  would  it  be  proper  to 
speak  of  the  former  in  a  work  on  the  latter.  In  a  framed 
girder,  the  lower  chord  and  those  diagonals  which  tend 
downwards  towards  the  middle  of  the  girder  are  subject  to 
tension.  The  better  material  to  resist  this  strain  is  wrought- 
iron,  and  this,  in  the  diagonals  at  least,  is  usually  employed. 
The  weight  with  which  this  material  may  be  safely  trusted 
per  square  inch  of  sectional  area  varies  according  to  the 
quality  of  the  metal,  from  7000  to  15,000  pounds.  Ordi- 
narily, it  may  be  taken  at  9000  pounds,  but  when  the  metal 


444  FRAMED    GIRDERS.  CHAP.    XXII. 

and  the  work  upon  it  are  of  superior  quality,  it  is  taken  at  as 
much  as  12,000,  or  even  higher  in  some  special  cases  This 
is  the  safe  power  of  the  metal  per  square  inch  of  the  sectional 
area.  Let  k  equal  this  power,  W  equal  the  load  to  be 
carried,  and  A  the  sectional  area  of  the  bar,  then 

Ak=  W 

W 
A  =  -  (298.) 

K 

As  an  application  of  this  formula,  take  the  case  of  the  di- 
agonal AP,  Fig.  93  ;  the  strain  in  which  is  29,896^  pounds. 
Putting  k  =  9000,  we  have 

29896! 

A  =  -        -^  =  3-3218 
9000 

or  the  rod  should  contain  3^-  inches  in  its  sectional  area. 
Referring  to  a  table  of  areas  of  circles,  we  find  that  the  rod, 
if  round,  should  be  a  trifle  over  2  inches  in  diameter,  or,  if 
a  flat  bar  4  inches  wide,  it  would  need  to  be  -J-  of  an  inch 
thick,  since  4x1  =  3-333. 

^  The  above  is  for  the  diagonals.  The  chords  are  usually 
of  wood.  When  so  made,  the  value  per  square  inch  sec- 
tional area  may  be  taken  at.one  tenth  of  the  ultimate  tensional 
power  of  the  materials  as  given  in  Table  XX.  Since  a  chord 
is  usually  compounded  of  three  or  more  pieces  in  width,  and 
of  lengths  less  than  the  length  of  the  chord,  it  is  necessary 
to  see  that  the  area  of  material  determined  by  the  use  of  for- 
mula (298.)  is  that  of  the  uncut  material,  or  of  the  uncut  sec- 
tional area  at  all  points  in  the  length.  Thus,  were  the  pieces 
so  assembled  as  to  have  no  two  heading  joints  occur  at  the 
same  point  in  the  length,  or  so  near  each  other  that  the  re- 
quisite bolts  for  binding  the  pieces  together  could  not  be  in- 
troduced between  the  two  joints,  then  the  uncut  sectional 


TENSION   IN   LOWER   CHORD.  445 

area  would  be  equal  to  that  of  all  the  pieces  in  the  width  ex- 
cept one.  Should  two  joints  occur  at  or  near  one  point  in 
the  length,  then  the  sectional  area  of  all  but  two  pieces  in 
width  must  be  taken  ;  and  so  on  for  other  cases. 

Where  care  is  exercised  in  locating  the  joints,  the  allow- 
ance for  joints,  bolt  holes,  and  other  damaging  contingencies 
may  be  taken  as  amounting  to  as  much  as  the  net  size  ;  or, 
ordinarily,  the  net  size  should  be  doubled.  Then  for  the 
total  sectional  area  we  have 


. 

IO  X  2         2O 

~^r\ 

20 


(299.) 


in  which  T  is  the  ultimate  resistance  to  tension,  as  found 
in  Table  XX. 

As  an  illustration,  take  the  case  of  the  lower  chord  of  Fig. 
93,  which,  at  the  middle  bay,  has  a  horizontal  strain  of  88,889 
pounds.  From  Table  XX.  we  have  the  resistance  to  tension 
of  Georgia  pine  equal  to  16,000  pounds.  By  formula  (299.) 

2oW      20x88880 

A  =  —~-  —  -  -~  =in    inches 

T  16000 

or  the  area  should  be  not  less  than  1  1  1  inches.  The  chord 
may  be  10  x  12  —  120  inches,  and  may  be  compounded  of 
three  pieces  in  width  —  a  centre  one  of  4x  12  and  two  out- 
side pieces  of  3x12  each. 


643. — Resistance  to  Compression. — The  top  chord  of  a 
framed  girder,  and  the  struts  or  diagonals   directed  down- 


44-6  FRAMED    GIRDERS.  CHAP.    XXII. 

ward  towards  the  points  of  support,  are  in  a  state  of  com- 
pression. 

The  rules  for  determining-  the  resistance  to  compression 
in  posts  or  struts  are  numerous,  and  their  discussion  has  oc- 
cupied many  minds.  The  theory  of  the  subject  will  not  be 
rehearsed  here.  For  this  the  reader  is  referred  to  authors 
who  have  made  it  a  special  point,  such  as  Tredgold,  Hodg- 
kinson,  Rankine,  Baker,  Francis  and  others. 

For  short  columns,  the  resistance  is,  approximately,  in 
proportion  to  the  area  of  cross-section  of  the  post.  As  the 
post  increases  in  length,  the  resistance  per  square  inch  of 
cross-section  gradually  diminishes. 

In  framed  girders,  the  struts,  and  also  the  chords,  when 
properly  braced  against  lateral  motion,  are  in  lengths  com- 
paratively short,  and  hence  the  resistance  which  the  material 
in  them  offers  is  not  much  less  than  when  in  short  blocks. 
Baker*  gives  as  the  strain  upon  posts 


=<:(< 


L* 


Reducing  this  expression  and  changing  the  symbols  to  agree 
with  those  of  this  work,  we  have 


in  which  /  equals  the  ultimate  resistance  of  the  post  per 
inch  of  sectional  area,  C  equals  the  ultimate  resistance  to 
compression  of  the  material  when  in  a  short  block,  e  the 
extension  of  the  material  per  foot  due  to  flexure,  within 

*  Strength  of  Beams,   Columns  and  Arches,  by  B.  Baker,  London,  1870, 
p.  182. 


RESISTANCE   TO    COMPRESSION.  447 

the  limits  of  elasticity,  as  found  in  Table  XX.,  and  d  is  the 
dimension  in  the  direction  of  the  bending.  This  in  a  post 
will  be  the  smaller  of  the  two,  or  the  thickness.  Let  h  rep- 
resent this  thickness  and  be  substituted  in  the  above  for  d\ 

then    r  =  j    is  the  ratio  of  the  length  to  the  thickness  or 
It 

smallest  dimension  of  the  cross-section  ;  /  and  h  both 
being  taken  of  the  same  denomination,  either  inches  or  feet. 
The  safe  limit  of  load  for  posts  is  variously  estimated  at 
from  6  to  10.  Putting  a  to  represent  this,  and  taking 
C  for  the  ultimate  resistance,  as  in  Table  XX.,  we  have  for 
the  safe  resistance 

f  —  __  £  _  (300.) 

- 


and  when     W   equals  the  load  to  be  carried,  and    A    equals 
the  sectional  area,  we  have 

W 
Af=  W     or      A  =  -j 

and,  by  substituting  for    /)     its  value,  as  in  (300.), 

W 


A  = 


C 


(S01} 


As  an  application,  let  it  be  required  to  find  the  area  of 
the  Georgia  pine  strut  AQ  in  Fig.  93,  the  strain  in  which 
is  (Art.  638),  say  45,000,  and  the  length  of  which  is  6-73. 

The  ratio  r  can  not  be  assigned  definitely  in  the  formula, 
as  h  is  unknown.  From  experience,  however,  a  value 
may  be  assigned  it  approximating  its  true  value,  and  after 
computation,  if  the  result  shows  that  the  assigned  value 
deviates  materially  from  the  true  value,  then  a  nearer  ap- 


443  FRAMED    GIRDERS.  CHAP.    XXII. 

proximation  may  be  made  for  a  second  computation.  The 
ratio  in  the  case  now  considered  is  probably  about  equal  to 
12.  We  will  take  it  at  this  amount  for  a  trial.  Take,  from 
Table  XX.,  the  values  of  6^=9500  and  e  =  0-00109,  for 
Georgia  pine.  Make  a,  the  factor  of  safety,  equal  to  10. 
The  value  of  W  is  45,000.  Then,  by  formula  (301.), 

10  [i  +(f  x  0-00109  x  I22)]  x  45000 
A  =  -  -  =  58-521 

9500 

or  the  area  should  be     58^     inches. 

Having  taken  the  ratio  at  12  we  should  have  the  thick- 
ness in  inches  equal  to  the  length  in  feet,  or  6-73.  Divid- 
ing the  area  58-521  by  this  gives  a  quotient  of  8-696  as 
the  breadth.  The  dimensions  of  the  piece  are  6f  x  8f .  If 
it  be  desirable  to  have  the  thickness  greater  than  here  given, 
then  a  second  trial  may  be  had  with  a  less  ratio. 

644. — Top     Chord      and     Diagonals— Dimensions. — By 

transformation  of  formula  (301.)  a  rule  may  be  arrived  at 
which  shall  define  the  breadth  of  a  diagonal  or  post  exactly. 
Let  A  =  hb,  and  let  //,  the  thickness,  bear  a  certain 
relation  to  b,  the  breadth  ;  or  nh  =  b,  n  being  a  con- 
stant assumed,  at  will  (for  example,  if  n  =  1-2,  then 
i-2/i  —  b).  Then  A  =  nh2.  Putting  also  for  r  its  value 

TO/ 

(/  being  taken  in  feet)  we  have 


h2n  = 


C 
Ctfn  ±=  Wa  + 


Ctfn  =  Wak3  +  (|  x  I22  Wael*} 
Ctfn  -  Wah3  =  |  x  12'  Wael* 


_ 
Cn  2          C 


TOP   CHORDS   AND   DIAGONALS.  449 

Completing  the  square  and  reducing  gives 


Cn      \       2Cn)  2Cn 


Let      W—~-     be  called     G ;     then  we  have 
2Cn 


2Cn 
and  by  substitution  the  above  formula  becomes 


h  =       V432Gel'+G'  +  G 


which  is  a  rule  to  ascertain  the  thickness  or  smallest  diame- 
ter of  a  strut  or  post,  and  in  which  /  is  in  feet  and  the 
other  dimensions  are  in  inches. 

This  rule,  owing  to  its  complication,  will  be  found  to  be 
tedious  in  practice.  For  this  reason,  formula  (301.)  ordi- 
narily, for  its  greater  simplicity,  is  to  be  preferred  ;  although, 
from  the  necessity  of  assuming  the  value  of  r,  a  second 
computation  may  be  required. 


645. — Example.  — What  is  the  value  of  h,  the  thickness 
of  the  strut  at  AQ,  *Fig.  93  ;  the  length  being  6-73,  and 
the  force  pressing  in  the  line  of  its  axis  being  45,000 
pounds. 

Putting  10  for  a,  the  factor  of  safety,  putting  1-2 
for  n,  the  factor  defining  the  relation  of  the  breadth  to  the 
thickness,  and  taking  from  Table  XX.  the  values  of  the  con- 
stants C  and  e  for  Georgia  pine,  we  have  £F=  45000, 
a  =  10,  *?  — 0-00109,  /=6-/3,  C  =9500  and  #=i-2. 


450  FRAMED    GIRDERS.  CHAP.    XXII. 

By  formula  (302.)  we  have 

r  a  45000  x  10 

G  =  W—^-  —  —  =  19-737 

2Cn      2  x  9500  x  i  -2 

Then,  by  formula  (303.  \ 


x  19-737  x  0-00109  x    )  4-  19-737  +  19-737 

=  6'943 

or  the  thickness  of  the  strut  is  required  to  be,  say  7  inches. 
As  nh  =  b,  therefore 

/;  =   I  -2  X  6-943  =  8-332 

equals  the  breadth  of  the  strut  ;  and  since  hb  =  A,  there- 
fore 

A  =  6-943x8-332  =  57-849 

equals  the  area  of  the  strut  ;  a  fraction  less  than  was  before 
found  by  formula  (301.).  That  value  would  have  been  the 
same  as  this  had  the  value  of  r  been  correctly  assumed. 
Its  exact  value  is  11-632  instead  of  12,  the  amount  there 
taken. 

64-6.  —  Derangement  from  Shrinkage  of  Timber*.  —  Ow- 

ing to  the  natural  shrinkage  of  timber  in  seasoning,  the  most 
carefully  framed  girder  will  settle  or  sag  more  or  less,  pro- 
vided adequate  measures  are  not  taken  to  prevent  it.  The 
ends  of  the  struts  press  upon  the  inside  of  the  chords,  while 
the  iron  rods  have  their  bearing  at  the  outside.  The  conse- 
quent diminution  in  height  of  the  girder  will  be  equal  to  the 
shrinkage  of  both  the  top  and  bottom  chords,  and  the  rods 
which  at  first  were  of  the  proper  length  will  be  found  cor- 
respondingly long. 

By  screwing  up  the  nuts  upon  the  rods  as  the  shrinkage 
progresses,  the  sagging  may  be  prevented  ;  but  this  would 
be  inconvenient  in  most  cases.  It  is  better,  in  constructing 
the  girder,  to  provide  bearings  of  metal  extending  through 


UNEQUAL   LOADS,    IRREGULARLY   PLACED. 


451 


the  depth  of  each  chord,  and  so  shaped  that  the  strut  and  rod 
shall  each  have  its  bearing  upon  it.  The  shrinkage  will  then 
have  no  effect  upon  the  integrity  of  the  frame. 


64-7. — Framed  Girder  with  Unequal  Loads  Irregularly 
Placed. — Let  Fig.  94  represent  such  a  case,  wherein  A,  B 
and  C  are  the  loads  upon  the  top  chord,  and  D  and 
E  the  loads  on  the  bottom  chord,  all  located  as  shown.  As 


FIG.  94. 

in  other  cases,  the  first  requirement  is  to  know  the  reactions 
at  the  two  supports  R  and  P.  In  a  girder  symmetrically 
loaded  this  involves  but  little  trouble,  as  the  half  of  the  total 
load  equals  the  reaction  at  each  support.  In  our  present 
case,  we  can  not  thus  divide  the  load,  since  the  reactions  are 
not  equal.  To  obtain  the  required  division  of  the  total  load, 
we  must  consider  each  of  the  several  weights  separately, 
dividing  it  between  the  two  supports  according  to  its  dis- 
tances from  them.  Thus,  putting  m  and  n  for  the  dis- 
tances of  the  load  A  from  the  two  supports,  the  portion  of 
A  bearing  upon  R  is  shown  by  formula  (#.)  (placing  A 
for  W), 


in  which     A,     the  weight,  is  multiplied  by     n,     its  distance 


452  FRAMED   GIRDERS.  CHAP.   XXII. 

from  the  opposite  support,  and  divided  by  /,  the  length  or 
span.  In  like  manner,  each  of  the  other  weights  may  be 
divided,  and  the  portion  bearing  upon  each  support  found. 

Putting  the  letters  o,  p,  q,  r,  s,  t,  n  and  v  to  repre- 
sent the  distances  shown  in  the  figure,  we  have,  as  the  total 
effect  upon  one  of  the  supports, 

An      Bp     Cr     Dv     Et 

*  =  —  +  7-  +  T  +  T  +  T 


R  = 


and  for  the  total  effect  upon  the  other  support, 

_  A  m  +  Bo  +  Cq  •+  Du  +  Es  •          (SO  5  ^ 

Adding  these  two  formulas,  we  have  as  the  total  effect  upon 
both  supports, 


A(m  +  ri)  +  B(o+p)+C(q  +  r)  +  D(u  +  v)  +  E(s  +  f) 

./v.  ~T~  Jr  — 


Here  the  sum  of  the  two  quantities  within  each  parenthesis 
is  equal  to     /    the  length,  and  consequently 


_ 

J\.-\-  r  —  ~ 


/ 

R+P  = A+B+C+Di E 

or  the  sum  of  the  reactions  of  the  two  supports  is  equal  to 
the  sum  of  all  the  weights.  In  this  we  have  proof  of  the 
accuracy  of  the  two  formulas  (304.)  and  (305.). 

648. — Load  upon  Each   Support— Graphical  Represen- 
tation.—The  value  of    R    in  formula  (304*)  may  be  readily 


DIVIDING  THE   LOADS   BETWEEN    SUPPORTS.  453 

found,  either  arithmetically  or  graphically.     The  formula  for 
one  weight,    R<  =  Aj     (d),  gives     R,l—An,    or  two  equal 

rectangles.    Having  three  of  these  quantities,    /,    A    and    n, 
the  fourth  quantity,     R»     may  be  graphically  found  thus  : 

In  Fig.  96  let    AB,    by  any  convenient  scale,  equal     n. 
Draw     AC    at  any  angle  with     AB,     and  equal  in  length  to 


FIG.  96. 

/.  Lay  off  AD  equal  to  A.  Join  B  with  C,  and  from 
D  draw  DE  parallel  with  CB.  AE  will  equal  Rt  the 
required  quantity,  for,  from  similar  triangles,  we  have 

AC  :  AB  ::  AD  :  AE 

I     :     n     :  :     A    :    R,  =  Aj 

To  obtain  the  value  of  R  for  all  of  the  weights,  proceed  as 
in  Fig.  97,  in  which  the  parallel  lines  FL,  GM,  HN,  JO 


H  J        S    l<  R  U    TCL 

FIG.  97. 


w 


and     KP    are  each  equal  to     /,     the  span    RP    of  Fig.  94. 


454 


FRAMED    GIRDERS. 


CHAP.    XXII. 


From  F  lay  off  upon  FL,  the  first  of  these  lines,  the 
distance  FV  equal  by  scale  to  the  weight  A  of  Fig.  94, 
and  from  F  on  line  FW  place  FQ  equal  to  n.  Con- 
nect Q  with  L.  From  V  draw  VG  parallel  with 
LQ.  FG  will  represent  R,. 

From  G  draw  GM  parallel  with  FL.  Make  GV 
equal  to  the  weight  B  (94),  and  GR  equal  to  /.  Con- 
nect R  with  M,  and  from  V  draw  VH  parallel  with 
MR.  GH  will  represent  Rs. 


H  J 


S   K  R  U    TO. 

FIG.   97. 


From  H  draw  //N  parallel  with  FL.  Make 
equal  to  the  weight  C  (Fig.  94),  and  HS  equal  to  r. 
Connect  5  with  N,  and  parallel  with  NS  draw  VJ. 
HJ  will  represent  R3. 

In  like  manner,  with  the  weight  D  and  distance  v  of 
Fig.  94,  obtain  ^A"  equal  to  Rt  ;  and  with  the  weight  E 
and  distance  t  obtain  KU  equal  to  R5. 

We  now  have  the  line    FU   equal  to  the  sum  of 


equals  that  portion  of  the  total  load  on  the  girder  which 
presses  upon  the  support    R. 


IRREGULAR   FORCE   DIAGRAM. 


455 


Similarly,  the  amount  of  pressure  upon  the  support  P 
may  be  obtained.  The  two,  R  and  P,  should  together 
equal  the  sum  of  the  weights  A,  B,  C,  D  and  E. 


649. — Girder    Irregularly    Loaded— Force     Diagram.— 

Having  accomplished  the  division  of  the  total   weight,  we 


FIG.  94. 


H 


E 

FIG.  95. 


may  now  construct  -upon  the  same  scale  with  that  of  Fig.  97, 
the  force  diagram,  Fig.  95,  for  the  girder  represented  in  Fig. 
94  and  described  in  Art,  647.  On  a  vertical,  RP,  make 


456  FRAMED   GIRDERS.  CHAP.   XXII. 

RM  equal  to  FU  (97);  and  RS  equal  to  the  weight  A, 
57'  equal  to  the  weight  B,  and  TP  equal  to  the  weight 
C,  all  as  in  Fig.  94.  Now,  since  RM  equals  FU  (97), 
equals  the  reaction  of  the  support  R.  therefore,  from  R 
draw  RE  parallel  with  RE  (94),  and  from  M  draw  ME 
parallel  with  ME  (94).  From  M  make  ML  equal  to  the 
weight  E  (94),  and  LK  equal  to  the  weight  D  (94). 
Draw  the  other  lines  all  parallel  with  the  corresponding 
lines  of  Fig.  94,  as  per  Arts.  618  and  619,  and  the  force  dia- 
gram will  be  complete. 

650. — Load  upon  Each  Support,  Arithmetically  Obtained. 

—The  reaction  of  the  two  supports  may  be  found  arithmet- 
ically, as  before  stated,  by  the  use  of  formulas  (304)  and 
(305.).  Thus,  let  the  several  weights  A,  B,  C,  D  and 
E  of  Fig.  94  be  rated,  by  the  scale  of  the  diagram,  at  15, 
23,  17,  22  and  19  parts  respectively.  These  parts  may 
represent  hundreds  or  thousands  of  pounds,  or  any  other 
denomination  at  will.  Let  /,  the  span,  equal  64,  and  the 
several  distances  n,  /,  r,  v  and  t  measure  respectively 
54,  34,  8,  26  and  44  by  the  same  scale. 
Formula  (304)  now  gives 

(15.  x  54)  +  (23  x  34)  +  (17  *  8)  +  (22  x  26)  +  (19  x  44)     . 

~6^~  =  49 

Formula  (305)  gives 

p  =  05  x  io)  +  (23x3o)  +  (i;x  56)  +  (22x  38) +  (19x20)  _ 

64 

R  -r  P  —  49  +  47  =  96 

The  sum  of  the  weights  is 

W—  15  +  23  +  17  +  22+19  =  96 

the  same  amount,  thus  proving  the  above  computation  cor- 
rect. 


QUESTIONS   FOR    PRACTICE. 


651. — Given  a  frame  similar  to  Fig.  87,  with  a  span  of  40 
feet,  a  height  of  23  feet,  with  the  length  of  the  vertical  BC 
equal  to  15  feet,  and  with  AF  and  BG  equal.  Draw  a 
diagram  of  forces,  and  show  what  the  strains  are  in  each  line 
of  the  frame;  the  three  loads  FG,  GH  and  HJ  being 
each  equal  to  5000  pounds. 

652. — According  to  the  rule  given  in  Art.  624,  show 
what  should  be  the  height  of  a  framed  girder  which  is  75 
feet  between  bearings. 

653. — According  to  Art.  627,  show  how  many  bays  the 
girder  of  the  last  article  should  have. 

654. — Show,  by  the  diagram  of  forces,  what  are  the 
strains  in  the  several  lines  of  a  girder  55  feet  long  between 
centres  of  bearings  and  5-27  feet  high  between  axes  of 
chords ;  the  girder  to  be  divided  into  five  equal  bays,  each 
being  an  isosceles  triangle  as  in  Fig.  93.  The  load  upon  the 
apex  of  each  triangle  is  5000  pounds,  and  that  suspended 
from  the  lower  chord  at  each  point  of  intersection  with  the 
diagonals  is  1250  pounds.  Letter  the  girder  as  in  Fig.  91. 

655.— To  test  the  accuracy  of  the  results  obtained  in  the 
last  article  compute  the  strains  arithmetically. 


FRAMED    GIRDERS.  CHAP.    XXII. 

656. — What  should  be  the  areas  of  cross-section  of  the 
bottom  chord  of  the  girder  of  Art.  654,  at  the  several  bays? 

What  should  be  the  sizes  of  the  upper  chord  and  of  the 
diagonal  struts?  The  timber  is  to  be  of  spruce  ;  a,  the  fac- 
tor of  safety,  to  be  taken  at  10,  and  n  at  1-2. 

What  should  be  the  areas  of  cross-section  of  the  diagonal 
rods,  taking  the  safe  strength  of  the  metal  at  9000  pounds  ? 

In  the  questions  of  this  Art.  take  the  strains  given  by  the 
diagram  of  forces. 


CHAPTER   XXIII. 

ROOF   TRUSSES. 

ART.  657. — Roof  Trusses  considered  as  Framed  Girders. 

—It  is  proposed,  in  this  chapter,  to  discuss  the  subject  of 
roof  trusses  in  so  far  only  as  they  may  be  considered  to  be 
framed  girders,  placed  in  position  to  carry  the  roofing  mate-' 
rial.  A  full  treatise  on  roofs  would  include  matter  extending 
beyond  the  limits  of  a  work  on  the  transverse  strain.  Those 
desirous  of  pursuing  the  subject  farther  are  referred  to  Tred- 
gold,  Bow  and  others*  who  have  written  more  fully  on 
roofs. 


658. — Comparison  of  Roof  Trusses. — Designs  for  roof 
trusses,  illustrating  various  principles  of  roof  construction, 
are  herewith  presented. 

The  designs  at  Figs.  98  to  102  are  distinguished  from  those 
at  Figs.  103  to  106,  by  having  a  horizontal  tie-beam.  In  the  lat- 
ter group,  and  in  all  designs  similarly  destitute  of  the  horizon- 
tal tie  at  the  foot  of  the  rafters,  the  strains  are  much  greater 
than  in  those  having  the  tie,  unless  the  truss  be  protected 
by  exterior  resistance,  such  as  may  be  afforded  by  competent 
buttresses. 

To  the  uninitiated  it  may  appear  preferable,  in  Fig;  103, 
to  extend  the  inclined  ties  to  the  rafters,  as  shown  by  the 
dotted  lines.  But  this  would  not  be  beneficial :  on  the  con- 

*  Tredgold's  Carpentry.     Bow's  Economics  of  Construction. 


460 


ROOF  TRUSSES. 


CHAP.    XXIII. 


trary,  it  would  be  injurious.  The  point  of  the  rafter  where 
the  tie  would  be  attached  is  near  the  middle  of  its  length, 
and  consequently  is  a  point  the  least  capable  of  resisting 
transverse  strains.  The  weight  of  the  roofing  itself  tends  to 
bend  the  rafter ;  and  the  inclined  tie,  were  it  attached  to  the 
rafter,  would,  by  its  tension,  have  a  tendency  to  increase  this 
bending.  As  a  necessary  consequence,  the  feet  of  the  rafters 
would  separate,  and  the  ridge  descend. 


98. 


99- 


100. 


K 


101. 


102. 


103. 


104. 


105. 


106. 


In  Fig.  104  the  inclined  ties  are  extended  to  the  rafters; 
but  here  the  horizontal  strut  or  straining  beam,  located  at 
the  points  of  contact  between  the  ties  and  rafters,  counteracts 
the  bending  tendency  of  the  rafters  and  renders  these  points 
stable.  In  this  design,  therefore,  and  only  in  such  designs,  is 
it  permissible  to  extend  the  ties  through  to  the  rafters. 
Even  here  it  is  not  advisable  to  do  so,  because  of  the  in- 
creased strain  produced.  (See  Figs.  118  and  120.)  The  design 
in  Fig.  103,  105  or  106  is  to  be  preferred  to  that  in  Fig.  104. 


LOAD    UPON   EACH    SUPPORT. 


461 


659- — Force  Diagram — Load  upon  Each  Support.- — By  a 

comparison  of  the  force  diagrams  hereinafter  given,  of  each 
of  the  foregoing  designs,  we  may  see  that  the  strains  in  the 
trusses  without  horizontal  tie-beams  at  the  feet  of  the  rafters 
are  greatly  in  excess  of  those  having  the  tie.  In  constructing 
these  diagrams,  the  first  step  is  to  ascertain  the  reaction  of, 
or  load  carried  by,  each  of  the  supports  at  the  ends  of  the 
truss.  In  symmetrically  loaded  trusses,  the  weight  upon  each 
support  is  always  just  one  half  of  the  whole  load. 


660. — Force  Diagram  for  Tru§s  in  Fig.  98. — To  obtain  the 
force  diagram  appropriate  to  the  design  in  Fig.  98,  first  letter 
the  figure  as  directed  in  Art.  619,  and  as  in  Fig.  107.  Then 


G- 


FIG.  108. 

draw  a  vertical  line,  EF  (Fig.  108),  equal  to  the  weight  W 
at  the  apex  of  the  roof;  or  (which  is  the  same  thing  in  effect) 
equal  to  the  sum  of  the  two  loads  of  the  roof,  one  extending 
on  each  side  of  W  half-way  to  the  foot  of  the  rafter.  Di- 
vide EF  into  two  equal  parts  at  G.  Make  GC  and 
GD  each  equal  to  one  half  of  the  weight  N.  Now,  since 
EG  is  equal  to  one  half  of  the  upper  load,  and  GD  to  one 
half  of  the  lower  load,  therefore  their  sum,  EG  +  GD  =  ED, 
is  equal  to  one  half  of  the  total  load,  or  to  the  reaction  of 
each  support,  E  or  F.  From  D  draw  DA  parallel 
with  DA  of  Fig.  107,  and  from  E  draw  EA  parallel  with 
EA  of  Fig.  107.  The  three  lines  of  the  triangle  AED  rep- 


462 


ROOF   TRUSSES. 


CHAP.    XXIII. 


resent  the  strains,  respectively,  in  the  three  lines  converging 
at  the  point  ADE  of  Fig.  i°7.  Draw  the  other  lines  of  the 
diagram  parallel  with  the  lines  of  Fig.  107,  and  as  directed  in 
Arts.  619  and  620.  The  various  lines  of  Fig.  108  will  repre- 
sent the  forces  in  the  corresponding  lines  of  Fig.  107;  bearing 
in  mind  (Art.  619)  that  while  a  line  in  the  forge  diagram  is 
designated  in  the  usual  manner  by  the  letters  at  the  two  ends 
of  it,  a  line  of  the  frame  diagram  is  designated  by  the  two 
letters  between  which  it  passes.  Thus,  the  horizontal  lines 
AD,  the  vertical  lines  AB,  and  the  inclined  lines  AE 
have  these  letters  at  their  ends  in  Fig.  108,  while  they  pass 
between  these  letters  in  Fig.  107. 

661. — Force  Diagram  for  Tru§§  in  Fig.  99. — For  this  truss 
we  have,  in  Fig.  109,  a  like  design,  repeated  and  lettered  as 


FIG.  109.  FIG.  no. 

required.  We  here  have  one  load  on  the  tie-beam  and  three 
loads  above  the  truss ;  one  on  each  rafter  and  one  at  the 
ridge.  In  the  force  diagram,  Fig.  no,  make  GH,  HJ  and 
JK,  by  any  convenient  scale,  equal,  respectively,  to  the 
weights  Gff,  HJ  and  JK  of  Fig.  109.  Divide  GK  into 
two  equal  parts  at  L.  Make  LE  and  LF  each  equal  to 
one  half  the  weight  EF  (Fig.  109).  Then  GF  is  equal  to 
one  half  the  total  load,  or  to  the  load  upon  the  support  G 
(Art.  660).  Complete  the  diagram  by  drawing  its  several 
lines  parallel  with  the  lines  of  Fig.  109,  as  indicated  by  the 
letters  (see  Art.  660),  commencing  with  GF,  the  load  on 


FORCE   DIAGRAMS. 


463 


the  support  G  (Fig.  109).  Draw  from  F  and  G  the  two 
lines  FA  and  GA,  parallel  with  these  lines  in  Fig.  109. 
Their  point  of  intersection  defines  the  point  A.  From  this' 
the  several  points  B,  C  and  D  are  developed,  and  the 
figure  completed.  Then  the  lines  in  Fig.  no  will  represent 
the  forces  in  the  corresponding  lines  of  Fig.  109,  as  indicated 
by  the  lettering.  (See  Art.  619.) 

662. — Force    Diagram  for  Truss  in  Fig.    100. — For   this 
truss  we  have,  in  Fig.  in,  a  similar  design,  properly  prepared 


B  D 

FIG.  112. 

by  weights  and  lettering  ;  and  in  Fig.  112  the  force  diagram 
appropriate  to  it. 


464 


ROOF   TRUSSES. 


CHAP.    XXIII. 


In  the  construction  of  this  diagram,  proceed  as  directed 
in  the  previous  example,  by  first  constructing  NS,  the  ver- 
tical line  of  weights ;  in  which  line  NO,  OP,  PQ,  QR  and 
RS  are  made  respectively  equal  to  the  several  weights 
above  the  truss  in  Fig.  in.  Then  divide  NS  into  two 


FIG.  ii2. 

equal  parts  at  T.  Make  TK  and  TL  each  equal  to  the 
half  of  the  weight  KL.  Make  JK  and  LM  equal  to  the 
weights  JK  and  LM  of  Fig.  in.  Now,  since  MN  is 
equal  to  one  half  of  the  weights  above  the  truss,  plus  one 
half  of  the  weights  below  the  truss,  or  half  of  the  whole 
weight,  it  is  therefore  the  weight  upon  the  support  N  (Fig. 


FORCE   DIAGRAMS. 


465 


in),  and  represents  the  reaction  of  that  support.  A  horizon- 
tal line  drawn  from  M  will  meet  the  inclined  line  drawn 
from  Nj  parallel  with  the  rafter  AN  (Fig.  in),  in  the 
point  At  and  the  three  sides  of  the  triangle  AMN  (Fig. 
112)  will  give  the  strains  in  the  three  corresponding  lines 
meeting  at  the  point  AMN  (Fig.  in).  The  sides  of  the  tri- 
angle HJS  (Fig.  ii2)  give  likewise  the  strains  in  the  three 
corresponding  lines  meeting  at  the  point  HJS  (Fig.  in). 
Continuing  the  construction,  draw  all  the  other  lines  of  the 
force  diagram  parallel  with  the  corresponding  lines  of  Fig. 
in,  and  as  directed  in  Art.  619.  The  completed  diagram 
will  measure  the  strains  in  all  the  lines  of  Fig.  in. 


663. — Force  Diagram  for  Truss  in  Fig.  101. — For  the  roof 
truss  at  Fig.  101  we  have,  in  Fig.  113,  a  repetition  of  it,  and  in 
Fig.  114  its  force  diagram. 


FIG.  113. 


FIG.  114. 


466  ROOF   TRUSSES.  CHAP.    XXIII. 

The  dimensions  on  the  vertical  line  HL  (Fig.  114)  are 
made  respectively  equal  to  the  weights  in  Fig.  113,  as  indi- 
cated by  the  lettering.  With  GH  equal  to  half  the  whole 
weight  on  the  truss  (Art.  660),  the  triangle  AGH  is  con- 
structed, giving  the  strains  in  the  three  lines  concentrating 
at  the  point  AGH  (Fig.  113).  Then,  drawing  the  other  lines 
parallel  with  the  corresponding  lines  of  Fig.  113,  the  com- 
pleted diagram  gives  the  strains  in  the  several  lines  of  that 
figure,  as  indicated  by  the  lettering.  (See  Art,  619.) 


664-. — Force  Diagram  for  Tru§s  in  Fig.  102. — The  roof 
truss  indicated  at  Fig.  102  is  repeated  in  Fig.  115,  with  the 
addition  of  the  lettering  required  for  the  construction  of  the 
force  diagram,  Fig.  116. 

In  this  case,  there  are  seven  weights,  or  loads,  above  the 
truss,  and  three  below.  Divide  the  vertical  line  OV  at 
Wy  into  two  equal  parts,  and  place  the  lower  loads  in  two 
equal  parts  on  each  side  of  W.  Owing  to  the  middle  one 
of  these  loads  not  being  on  the  tie-beam  with  the  other  two, 
but  on  the  upper  tie-beam,  the  line  GH,  its  representative 
in  the  force  diagram,  has  to  be  removed  to  the  vertical  BJ, 
and  the  letter  M  is  duplicated.  The  line  NO  equals  half 
the  whole  weight  of  the  truss,  or  3^  of  the  upper  loads,  plus 
one  of  the  lower  loads,  plus  half  of  the  load  at  .the  upper  tie- 
beam.  It  is  therefore  the  true  reaction  of  the  support  NO, 
and  AN  is  the  horizontal  strain  in  the  beam  there.  It  will 
be  observed  also,  that  while  HM  and  GM  (Fig.  116), 
which  are  equal  lines,  show  the  strain  in  the  lower  tie-beam 
at  the  middle  of  the  truss,  the  lines  CH  and  FG,  also 
equal  but  considerably  shorter  lines,  show  the  strains  in  the 
upper  tie-beam.  Ordinarily  in  a  truss  of  this  design,  the 
strain  in  the  upper  beam  would  be  equal  to  that  in  the  lower 
one,  which  becomes  true  when  the  rafters  and  braces  above 


FORCE   DIAGRAMS. 


467 


the  upper  beam  are  omitted.    In  the  present  case,  the  thrusts 
of  the  upper  rafters    produce   tension   in  the  upper  beam 


FIG.  115. 


FIG.  1 1 6. 


equal  to  CM  or  FM  of  Fig.  116,  and  thus,  by  counteract- 
ing the  compression  in  the  beam,  reduce  it  to  CH  or  FG 
of  the  force  diagram,  as  shown. 


468 


ROOF   TRUSSES, 


CHAP.   XXIII. 


665. — Force  Diagram  for  Truss  In  Fig.   103. The   force 

diagram  for  the  roof  truss  at  Fig.  103  is  given  in  Fig.  118, 
while  Fig.  117  is  the  truss  reproduced,  with  the  lettering 
requisite  for  the  construction  of  Fig.  118. 


FIG.  117. 


FIG.  118. 


The  vertical  EF  (Fig.  118)  represents  the  load  at  the 
ridge.  Divide  this  equally  at  W,  and  place  half  the  lower 
weight  each  side  of  W,  so  that  CD  equals  the  lower 
weight.  Then  ED  is  equal  to  half  the  whole  load,  and 
equal  to  the  reaction  of  the  support  E  (Fig.  117).  The  lines 
in  the  triangle  ADE  give  the  strains  in  the  corresponding 
lines  converging  at  the  point  ADE  of  Fig.  117.  The  other 
lines,  according  to  the  lettering,  give  the  strains  in  the  cor- 
responding lines  of  the  truss.  (See  Art.  619.) 


666. — Force  Diagram  for  Truss  in  Fig.  104. — This  truss  is 
reproduced  in  Fig.  119,  with  the  letters  proper  for  use  in  the 
force  diagram,  Fig.  120. 

Here  the  vertical  GK,  containing  the  three  upper  loads 
GH,  HJ  and  JK,  is  divided  equally  at  W,  and  the  lower 
load  EF  is  placed  half  on  each  side  of  W,  and  extends 
from  E  to  F.  Then  FG  represents  one  half  of  the 
whole  load  of  the  truss,  and  therefoie  the  reaction  of  the  sup- 
port G  (Fig.  119).  Drawing  the  several  lines  of  Fig.  120 
parallel  with  the  corresponding  lines  of  Fig.  119,  the  force 


FORCE   DIAGRAMS. 


469 


diagram  is  complete,  and  the  strains  in  the  several  lines  of 
119  are  measured  by  the  corresponding  lines  of  120,  (See 
Art.  619.) 


FIG,   120. 


A  comparison  of  the  force  diagram  of  the  truss  in  Fig.  117 
with  that  of  the  truss  in  Fig.  119  shows  much  greater  strains 
in  the  latter,  and  we  thus  see  that  Fig.  117,  or  103,  is  the  more 
economical  form. 


667. — Force  Diagram  for  Truss  in  Fig.  105. — This  truss 
is  reproduced  and  prepared  by  proper  lettering  in  Fig.  121, 
and  its  force  diagram  is  given  in  Fig.  122. 

Here  the  vertical  JM  contains  the  three  upper  loads 
JK,  KL  and  LM.  Divide  JM  into  two  equal  parts  at 


47° 


ROOF  TRUSSES. 


CHAP.    XXIII. 


G,  and  make  FG  and  GH  respectively  equal  to  the  two 
loads  FG  and  GH  of  Fig.  121.  Then  HJ  represents  one 
half  of  the  whole  weight  of  the  truss,  and  therefore  the  reac- 
tion of  the  support  J.  From  H  and  J  draw  lines  par- 
allel with  AH  and  AJ  of  Fig.  121,  and  the  sides  of  the  tri- 


FlG.    121. 


M 


FIG.  122. 


angle  AHJ  will  give  the  strains  in  the  three  lines  concen- 
trating in  the  point  AHJ  (Fig.  121).  The  other  lines  of  Fig. 
122  are  all  drawn  parallel  with  their  corresponding  lines  in 
Fig.  121,  as  indicated  by  the  lettering.  (See  Art.  619.) 


FORCE  DIAGRAMS. 


471 


FIG.  123. 


FIG.  124. 


472 


ROOF   TRUSSES. 


CHAP.   XXIII. 


668. — Force  Diagram  for  Truss  in  Fig.  106. — This  truss 
is  reproduced  in  Fig.  123  with  the  lettering  proper  for  its 
force  diagram,  as  given  in  Fig.  124.  The  five  external  weights 
of  Fig.  123  make  up  the  line  LQy  and  the  two  internal 
weights  are  set,  one  on  each  side  of  y,  the  middle  point  of 
LQj  extending  to  H  and  K.  KL  equals  one  half  the 
weight  of  the  whole  truss,  and  equals  the  reaction  of  the 
point  of  support  L  (Fig.  123).  The  sides  of  the  triangle 
AKL,  therefore,  give  the  respective  strains  in  the  three  lines 
converging  at  the  point  AKL  of  Fig.  123.  The  other  lines 
of  Fig.  124  are  found  in  the  usual  manner.  (See  Art.  619.) 

669.— Strains  in  Horizontal  and  Inclined  Ties  Compared. 

—A  comparison  between  a  truss  with  a  horizontal  tie  at  the 

£• 


FIG.  125. 

feet  of  the  rafters,  and  one  without  such  tie  will  now  be 
given.  The  truss  without  a  horizontal  tie  shown  in  Fig.  103 
is  one  of  the  simplest  in  construction,  and  is  suitable  for  the 
comparison.  Repeating  it  in  Fig.  125,  and  adding  the  dotted 
lines,  we  have  likewise  the  form  of  a  truss  with  a  horizontal 
tie.  From  Art.  608  we  have,  in  formula  (293.\  for  the  hori- 
zontal strain, 

H,  =  W± 

in  which  Wt  .equals  the  total  weight  of  the  truss  and  its 
load  (Fig.  125),  //  equals  half  the  span,  equals  AD,  and  c 


HORIZONTAL  AND    INCLINED    TIES.  473 

W 

equals  twice  the  height,  equals    2DE.       By  putting    P  —  — 

equals  the  reaction   of  one   of  the  supports    A    or    B,    and 
putting    d    for    DE,    we  have 


or,  from  Fig.  125, 

DE  :  AD  :  :  P  :  H 

d    :     h     :  :  P  :  H  =  P-, 

a 

that  is  to  say,  when  the  vertical  DE  represents  half  the 
weight  of  the  truss,  then  AD  may  be  put  to  represent  the 
horizontal  strain.  Draw  CF  horizontal,  and  by  similar  tri- 
angles we  have 

DE  :  AD  :  :  CE  :  CF  or 

CE   :    CF  ::    P    :    H  =  P^ 

LJtL 

or,  with  CE  put  to  represent  one  half  the  weight  of  the 
whole  truss,  then  CF,  by  the  same  scale,  will  measure  the 
horizontal  strain. 

Under  these  conditions,  CF  measures  the  horizontal 
strain  in  either  truss,  whether  with  or  without  a  tie-beam. 
If  the  truss  have  a  horizontal  tie  AB,  then  CF  measures 
the  tension  in  this  tie.  If  it  be  without  the  tie  AB,  having 
instead  thereof  the  raised  tie  ACB,  then  still  CF  mea- 
sures the  horizontal  strain  at  A  or  BJ  but  not  the  strain  in 
the  raised  tie  AC. 

The  strain  in  this  inclined  tie  is  measured  by  the  line 
AC,  for  the  three  sides  of  the  triangle  ACE  are  in  propor- 
tion as  the  strains  in  these  lines  respectively  (see  Art.  619), 
therefore  the  strains  in  the  ties  of  the  two  trusses  are  compa- 
rable by  the  two  lines  CF  and  AC. 


474 


ROOF   TRUSSES. 


CHAP.    XXIII. 


The  compressive  strain  in  the  rafter  is  also  correspond- 
ingly increased;  for  just  in  proportion  as  AE  exceeds 
£F,  so  does  the  compressive  strain  in  the  rafter  of  a  truss 
with  an  inclined  tie  exceed  that  of  one  with  a  horizontal  tie. 


670. — Vertical  Strain  in  Tru§§  with  Inclined    Tie. — In 

Fig.  125,  if  the  inclined  tie  were  lowered,  so  that  the  point   C, 


FIG.  125. 

descending,  should  reach  the  point  D ;  or,  if  the  inclined 
tie  become  the  horizontal  tie  AB ;  then  the  vertical  rod 
DE  would  be  subject  to  no  strain  from  the  weight  of  the 
rafters  and  the  load  upon  them.  In  the  absence  of  the 
horizontal  tie,  or  when  the  inclined  tie  is  depended  upon  to 
resist  the  spreading  of  the  rafters,  the  vertical  rod  CE  is 
strained  directly  in  proportion  to  CD,  the  elevation  of  the 
tie,  and  inversely  as  the  height  CE.  This  relation  may  be 
shown  as  follows : 

Let  P  be  put  for  DE  (Fig.  125)  and  represent  one  half 
the  weight  of  the  truss.  Then  AD  will  represent  the 
horizontal  strain  at  A  ;  or,  representing  the  span  AB  by 


then  equals     AD     equals  the  horizon- 


the  symbol 

tal  strain.     Putting    a    for    CD    and    d    for     DE     we  have 
the  proportion 


ENHANCED    STRAINS    FROM    INCLINED   TIES.  475 

DE  :  AD  : :  P  :  H 


or  d    :     -     ::  P'  H=P-^ 

2  2d 

and  also,  AD  :  CD    : :  H  :   V 


S-      :     a     :VJ5T:   V  =  H^  =  P 

2  s  d 


by  substitution,  or 


This  gives  the  vertical  strain  in  CE,  due  to  the  raising 
of  the  tie  from  D  to  C,  but  it  is  not  the  whole  of  the 
strain  ;  it  is  only  so  much  of  the  vertical  strain  as  is  due  to 
the  weight  of  the  roof.  The  tension  thus  found  in  CE  is 
sustained  at  E  by  the  two  rafters,  and,  passing  through 
them  to  A  and  /?,  creates  horizontal  and  vertical  thrusts 
precisely  as  did  the  original  weight.  The  vertical  tension 
thus  brought  to  CE  again  acts  as  a  weight  at  £,  and, 
passing  down  the  rafters  and  through  the  tie  back  to  C, 
again  adds  a  load  at  C.  This  in  turn  passes  around  and  re- 
turns to  C,  adding  to  the  load  ;  and  so  on  in  an  endless 
round  to  infinity.  But  the  successive  strains  thus  generated 
are  in  a  decreasing  series,  and  they  may  therefore  be  summed 
up  and  defined.  Thus,  as  has  just  been  shown,  the  vertical 
effect  from  the  weight  of  the  roof  is 


The  vertical  effect  of  this  latter  is 


d:  a  ::    V    :    V  =  V--^p 

d 


476  ROOF   TRUSSES.  CHAP.    XXIII. 

The  vertical  effect  of  this  is 

d  :  a  ::    V  :   V"  =  V'r  =  P\- 

d  \d 

The  next  term  in  the  series  will  be 


and  the  sum  of  all  the  terms  will  be 


a       /  a 


showing  that  the  several  values  of  the  fraction  by  which  the 
weight  P  is  multiplied  constitute  a  geometrical  series,  with 

—j-  for  the  first  term  and  -j-  for  the  ratio.  Since  —j-  is 
a  da 

less  than  unity,  we  have  a  geometrically  decreasing  infinite 
series,  the  sum  of  which  is  equal  to  the  first  term  divided  by 
one  minus  the  ratio,*  or 

a 

c-       ^  a 

a       d  —  a 


and,  since     d—a~b    ot  Fig.  125, 


We  have,  therefore,  as  the  total  vertical  effect  due  to  the 
elevation  of  the  middle  of  the  tie  from     D    to    C, 


Ray's  Algebra,  Part  Second,  Art.  299, 


INCLINED   TIES — ILLUSTRATIONS. 


477 


or  the  vertical  effect  is  directly  in  proportion  to  CD,  the 
elevation  of  the  tie,  and  inversely  in  proportion  to  CE,  the 
length  of  the  vertical  tie-rod. 

671. — Illustrations. — To  illustrate  the  effect  of  the  eleva- 
tion of  the  tie-rod,  upon  the  vertical  strain  in  the  suspension- 
rod,  let  the  point  C,  Fig.  125,  be  elevated  -J-  of  the  verti- 


Fio.   125. 
cal  height  of  the  truss  above  the  horizontal  line    AB.     Here 

a  =  i     and     b  ==  4,     and    -y-  =  \  ;     or 


When  the  elevation  equals    £    of  the  entire  height,  then 


When  the  elevation  equals    \    of  the  entire  height,  then 


When  the  elevation  equals    \    of  the  whole  height,  then 


Thus  it  is  seen,  in  this  last  case,  that  the  effect  due  to  the 


4/8  ROOF   TRUSSES.  CHAP.    XXIII. 

elevation  of  the  tie-beam  is  equal  to  that  of  doubling-  the 
whole  weight  of  the  roof,  and  this  increase  affects  not  only 
the  vertical  suspension  rod  at  the  middle,  but  also  the  rafters 
and  inclined  ties,  as  was  shown  at  Art.  669. 

When,  therefore,  in  order  to  gain  a  small  additional 
height  to  the  interior  of  a  building,  it  is  proposed  to  raise  the 
middle  point  of  the  tie-rod,  it  would  seem  advisable  to  con- 
siaer  whether  this  small  additional  height  be  an  adequate 
compensation  for  the  increased  strains  thereby  induced,  and 
the  consequent  enhanced  cost  for  material  necessary  to  re- 
sist these  strains  ;  and  also,  whether  it  be  not  more  advisable 
to  raise  the  walls  of  the  building,  rather  than  the  ties  of  the 
trusses. 


672. — Planning  a  Roof. — In  designing  a  roof  for  a  build- 
ing, the  first  point  requiring  attention  is  the  location  of  the 
trusses.'  These  should  be  so  placed  as  to  secure  solid  bear- 
ings upon  the  walls ;  care  being  taken  not  to  place  either  of 
the  trusses  over  an  opening,  such  as  those  for  windows  or 
doors,  in  the  wall  below.  Ordinarily,  trusses  are  placed  so 
as  to  be  centrally  over  the  piers  between  the  windows  ;  the 
number  of  windows  consequently  ruling  in  determining  the 
number  of  trusses  and  their  distances  from  centres.  This 
distance  should  be  from  ten  to  twenty  feet ;  fifteen  feet  apart 
being  a  suitable  medium  distance.  The  farther  apart  the 
trusses  are  placed,  the  more  they  will  have  to  carry  ;  not 
only  in  having  a  larger  surface  to  support,  but  also  in  that 
the  roof  timbers  will  be  heavier ;  for  the  size  and  weight  of 
the  roof  beams  will  increase  with  the  span  over  which  they 
have  to  reach. 

In  the  roof-covering,  itself,  the  roof-planking  may  be  laid 
upon  jack-rafters,  carried  by  purlins  supported  by  the 
trusses ;  or  upon  roof  beams  laid  directly  upon  the  back  of 


PLANNING   A    ROOF — LOAD    ON   TRUSS.  479 

the  principal  rafters  in  the  trusses.  In  either  case,  proper 
struts  should  be  provided,  and  set  at  proper  intervals  to  re- 
sist the  bending  of  the  rafter.  In  case  purlins  are  used,  one 
of  these  struts  should  be  placed  at  the  location  of  each 
purlin. 

The  number  of  these  points  of  support  rules  largely  in 
determining  th'e  design  for  the  truss,  thus  : 

For  a  short  span,  where  the  rafter  will  not  require  sup- 
port at  an  intermediate  point,  Fig.  98  or  103  will  be 
proper. 

For  a  span  in  which  the  rafter  requires  supporting  at  one 
intermediate  point,  take  Fig.  99,  104  or  105. 

For  a  span  with  two  intermediate  points  of  support  for 
the  rafter,  take  Fig.  100  or  106. 

For  a  span  with  three  intermediate  points,  take  Fig.  102. 

Generally,  it  is  found  convenient  to  locate  these  points  of 
support  at  nine  to  twelve  feet  apart.  They  should  be  suf- 
ficiently close  to  make  it  certain  that  the  rafter  will  not  be 
subject  to  the  possibility  of  bending. 


673 — Load  upon  Roof  Truss.— In  constructing  the  force 
diagram  for  any  truss,  it  is  requisite  to  determine  the  points 
of  the  truss  which  are  to  serve  as  points  of  support  (see 
Figs.  109,  in,  etc.),  and  to  ascertain  the  amount  of  strain,  or 
loading,  which  will  occur  at  every  such  point. 

The  points  of  support  along  the  rafters  will  be  required 
to  sustain  the  roofing  timbers,  the  planking,  the  slating,  the 
snow,  and  the  force  of  the  Avind.  The  points  along  the  tie- 
beam  will  have  to  sustain  the  weight  of  the  ceiling  and  the 
flooring  of  a  loft  within  the  roof,  if  there  be  one,  together 
with  the  loading  upon  this  floor.  The  weight  of  the  truss 
itself  must  be  added  to  the  weight  of  roof  and  ceiling. 


ROOF  TRUSSES.  CHAP.    XXIII. 

674. — Load  on  Roof  per  Foot  Horizontal. — In  any  im- 
portant work,  each  of  the  items  in  Art.  673  should  be  care- 
fully estimated,  in  making  up  the  load  to  be  carried.  For 
ordinary  roofs,  the  weights  may  be  taken  per  foot  superficial, 
as  follows  : 

Slate,  about     7-0     pounds. 

Roof  plank,  "         2-7  " 

Roof  beams,  or  jack-rafters,    "         2-3  " 

In  all,  12         pounds. 

This  is  for  the  superficial  foot  of  the  inclined  roof.  For  the 
foot  horizontal,  the  augmentation  of  load  due  to  the  angle  of 
the  roof  will  be  in  proportion  to  its  steepness.  In  ordinary 
cases,  the  twelve  pounds  of  the  inclined  surface  will  not  be 
far  from  fifteen  pounds  upon  the  horizontal  foot. 
For  the  roof  load  we  may  take  as  follows : 

Roofing,  about  15     pounds. 

Roof  truss,  "           5 

Snow,  "  20           " 

Wind,  "  10 

Total  on  roof,          50     pounds 

per  square  foot  horizontal. 

This  estimate  is  for  a  roof  of  moderate  inclination,  say 
one  in  which  the  height  does  not  exceed  i  of  the  span. 
Upon  a  steeper  roof,  the  snow  would  not  gather  so  heavily, 
but  the  wind,  on  the  contrary,  would  exert  a  greater  force. 
Again,  the  wind  acting  on  one  side  of  a  roof  may  drift  the 
snow  from  that  side,  and  perhaps  add  it  to  that  already 
lodged  upon  the  opposite  side.  These  two,  the  wind  and 
the  snow,  are  compensating  forces.  The  action  of  the  snow 
is  vertical :  that  of  the  wind  is  horizontal,  or  nearly  so.  The 
power  of  the  wind  in  this  latitude  is  not  more  than  thirty 


LOADING — SELECTION   OF   DESIGN.  481 

pounds  upon  a  superficial  foot  of  a  vertical  surface  ;  except, 
perhaps,  on  elevated  places,  as  mountain  tops  for  example, 
where  it  should  be  taken  as  high  as  fifty  pounds  per  foot  of 
vertical  surface. 


675. — Load  upon  Tie-Beam. — The  load  upon  the  tie- 
beam  must  of  course  be  estimated  according  to  the  require- 
ments of  each  case.  If  the  timber  is  to  be  exposed  to  view, 
the  load  to  be  carried  will  be  that  only  of  the  tie-beam  and 
the  timber  struts  resting  upon  it.  If  there  is  to  be  a  ceiling 
attached  to  the  tie-beam,  the  weight  to  be  added  will  be  in 
accordance  with  the  material  composing  the  ceiling.  If  of 
wood,  it  need  not  weigh  more  than  two  or  three  pounds  per 
foot.  If  of  lath  and  plaster,  it  will  weigh  about  nine  pounds ; 
and  if  of  iron,  from  ten  to  fifteen  pounds,  according  to  the 
thickness  of  the  metal.  Again,  if  there  is  to  be  a  loft  in  the 
roof,  the  requisite  flooring  may  be  taken  at  five  pounds,  and 
the  load  upon  the  floor  at  from  twenty-five  to  seventy 
pounds,  according  to  the  purpose  for  which  it  is  to  be  used. 

676.— Selection  of  Design  for  Roof  Tru§s. — As  an  ex- 
ample in  designing  a  roof  truss  :  Let  it  be  required  to  provide 
trusses  for  a  building  measuring  60  x  90  feet  to  the  centre 
of  thickness  of  the  walls,  with  seven  windows  upon  each 
side,  and  with  a  roof  having  its  height  equal  to  one  third  of 
the  span.  The  roofing  is  to  be  of  plank  and  slate,  the  ceil- 
ing is  to  be  finished  with  plastering,  and  the  space  within 
the  roof  is  to  be  used  for  the  storage  of  light  articles,  not  to 
exceed  twenty-five  pounds  to  the  square  foot. 

Here,  in  the  first  place,  we  have  to  determine  the  number 
of  trusses.  As  there  are  seven  windows  on  a  side,  there 
should  be  six  trusses,  one  upon  each  pier  between  the  win- 
dows. The  six  trusses  and  the  two  end  walls  will  afford 


4^2  ROOF   TRUSSES.  CHAP.    XXIII. 

eight  lines  of  support  for  the  roofing.  There  will  thus  be 
seven  bays  of  roofing  of  Sf-  =  \2-\  feet  each,  and  this  is  the 
width  of  roofing  to  be  carried  by  each  truss. 

In  the  next  place,  the  points  of  support  in  the  truss  are 
to  be  ascertained.  If  these  are  provided  at  every  ten  feet 
horizontally,  they  will  divide  the  half  truss  into  three  spaces, 
and  there  will  be  two  intermediate  points  of  support.  For 
this  arrangement,  such  a  roof  truss  as  is  shown  in  Fig.  100  will 
be  appropriate,  but  if  the  space  in  the  roof  is  required  to  be 
quite  unobstructed  with  timber  at  the  middle,  then  a  modifi- 
cation of  this  design  may  be  used,  as  in  the  form  shown  in 
Fig.  126 ;  each  rafter  being  still  divided  into  three  equal 
parts. 

677. — Load  on  Each   Supported    Point  in  Tru§§. — The 

horizontal  measurement,  then,  of  the  roofing  to  be  carried 
by  each  supported  point  in  the  truss,  will  be  10  feet  along 
the  line  of  the  truss  and  12-f-  feet  across  the  truss  (this 
latter  being  the  width  of  each  bay  as  above  found);  or 
lox  12-f-  =  128^-  feet.  With  a  weight  per  foot  of  50  pounds, 
as  estimated  in  Art.  674,  we  have,  for  the  load  upon  each 
supported  point  of  the  truss, 

1284-  X   50  =   64284 

or,  say    6500    pounds. 


678. — Load   on   Each  Supported  Point  in  Tie-Beam. — 

The  tie-beam  having  two  points  of  support,  we  have 
*£-  =  20  feet  for  the  length  of  the  surface  to  be  carried. 
This,  multiplied  by  the  width  between  trusses,  gives 
20  x  \2\  =  257^  feet  area  of  surface  to  be  carried  by  each 
point  of  support.  We  will  estimate  the  weight  per  foot  in 
this  present  case  as  follows: 


LOADING — CONSTRUCTING    FORCE    DIAGRAM.  483 

Load  upon  the  floor,  25     pounds. 
Flooring,  with  timber,  5 

Plastering,  9 

Tie-beam,  etc.,  i     pound. 

Total  at  tie-beam,      40     pounds. 

This  gives 

2571-  x  40  =  102856- 

or,  say     10,300     pounds  upon  each  supported  point. 

Therefore,  the  two  balls  GH  and  HJ,  suspended 
from  the  tie-beam  of  Fig.  126,  are  to  be  taken  as  weighing 
10,300  pounds  each,  while  the  five  balls  located  above  the 
rafters  are  to  be  understood  as  weighing  6500  pounds 
each  (Art.  677). 

679. — Contracting  the  Force  Diagram. — We   may  now 

proceed  to  construct  the  force  diagram,  Fig.  127,  as  follows: 
Upon  the  vertical  line  KP  lay  off  in  equal  parts  KL, 
LM,  MN,  NO  and  OP,  according  to  any  convenient 
scale,  each  equal  to  6500  pounds — the  weight  of  the  balls 
above  the  rafters  (Art.  677).  If  a  scale  of  100  parts  to  the 
inch  be  selected  for  the  force  diagram,  and  each  part  be 
understood  as  representing  100  pounds,  then  *££•£-  =  65, 
equals  the  number  of  parts  to  assign  to  each  of  the  distances 
KLj  LM,  etc.,  and  each  will  be  T6^5y  of  an  inch  in  length. 
Dividing  KP  at  H  into  two  equal  parts,  lay  off  on  each  side 
of  H  the  distances  GH  and  HJ^  each  equal,  by  the 
scale,  to  10,300  pounds.  This  distance  is  found  by  dividing 
10,300  by  IOD  ;  the  quotient  103  is  the  number  of  parts, 
and  the  distances  will  each  be  fjj-J,  or  one  inch  and 
of  an  inch  in  length.* 


*  The  scale  here  selected,  although  sufficient  for  the  purposes  of  illustration, 


484 


ROOF  TRUSSES. 


CHAP.    XXIII. 


HK  now  represents  one  half  the  weight  upon  the 
rafters,  and  HJ  one  half  the  load  upon  the  tie-beam,  and 
their  sum,  JK,  equals  one  half  the  total  load  of  the  truss, 
equals  the  load  upon  the  point  of  support  K. 


FIG.  126. 


FIG.  127. 

From     y,    H    and     G    draw  the  horizontal  lines     JA, 
HD     and     GF.     From     K,     L,     M,     N,     O     and     P    draw 


would  be  too  small  for  a  working  drawing.  For  the  latter,  a  scale  should  be 
selected  as  large  as  can  be  conveniently  used,  such  as  10  parts  to  the  inch, 
and  too  pounds  to  each  part.  This  would  give  1000  pounds  to  the  inch, 
and  each  of  the  distances  KL,  LM,  etc.,  would  measure  6^  inches. 

It  must  also  be  remembered  that  the  accuracy  of  the  force  diagram  depends 
upon  the  care  with  which  the  distances  upon  the  vertical  line  are  laid  off  and 
the  lines  drawn.  The  drawing  implements  should  be  examined  to  know  that 
they  are  true,  and  each  line  should  be  drawn  carefully  parallel  with  the  corre- 
sponding line  of  the  truss.  Unless  this  care  is  exercised,  the  results  may  differ 
considerably  from  the  truth. 


MEASURING   THE   STRAINS.  485 

lines,  as  shown,  carefully  parallel  with  the  rafters.  From 
F  and  A  draw  the  lines  FE  and  AB,  parallel  with 
the  two  braces.  Connect  B  and  E  by  the  vertical  line 
BE,  and  then  the  force  diagram  is  complete. 

680. — Uleasanrtiig  tlie  Force  Diagram. — After  drawing 
the  lines  of  the  diagram  as  above  directed,  they  should  all 
be  carefully  traced  to  know  that  the  required  conditions  are 
fulfilled,  or  that  each  set  of  lines,  drawn  parallel,  in  the  dia- 
gram of  forces,  to  the  lines  converging  to  a  point  in  the 
truss,  forms  a  closed  polygon.  (See  Arts.  618,  619  and  620.) 

The  diagram,  by  this  test,  having  been  found  correct,  the 
force  in  each  line  of  the  truss  may  be  measured  by  applying 
the  scale  to  the  corresponding  line  of  the  diagram. 

For  example,  take  the  strains  in  one  of  the  rafters.  At 
its  lower  end,  or  the  part  A K,  its  corresponding  line  AK 
of  Fig.  127  measures  478  parts,  by  the  same  scale  with 
which  the  weights  on  the  vertical  line  KP  were  laid  off. 
This,  at  100  pounds  to  the  part,  gives  47,800  pounds  as 
the  strain  in  the  foot  of  the  rafter.  The  next  section  of  the 
rafter  is  designated  by  the  letters  BL,  and  the  line  BL 
(Fig.  127)  measures  420,  and  indicates  a  strain  in  this  part  of 
the  rafter  of  42,000  pounds.  The  third  or  upper  portion 
of  the  rafter  is  designated  by  the  letters  CM,  and  the  cor- 
responding line  in  Fig.  127  measures  58  parts,  indicating 
5800  pounds  as  the  strain  in  the  upper  end  of  the  rafter. 

For  the  brace  AB  we  have  the  line  AB  (Fig.  127), 
measuring  58  parts  of  the  scale,  and  indicating  5800 
pounds  as  the  strain  in  the  brace. 

For  the  vertical  BD  we  have  the  line  BD  (Fig.  127) 
measuring  135  parts  of  the  scale,  and  indicating  13,503 
pounds  as  the  strain  in  the  vertical. 

For  the  horizontal  strains,  we  have  for  CD,  the  corre- 
sponding line  in  Fig.  127,  which  measures  301  parts,  and 


486  ROOF   TRUSSES.  CHAP.    XXIII. 

gives  30,100  pounds  as  the  strain.  For  DH,  the  middle 
portion  of  the  tie-beam,  DH  (Fig.  127)  measures  350, 
showing  the  strain  to  be  35,000  pounds;  and  lor  AJ,  or 
one  end  of  the  tie-beam,  AJ  (Fig.  127)  measures  398 
parts,  and  gives  39,800  pounds  as  the  strain. 

The  strains  in  the  other  and  corresponding  parts  of  the 
truss  are  the  same  as  these,  so  that  we  now  have  all  the 
strains  required. 

681. — Strains  Computed  Arithmetically. — Instead  of  de- 
pending solely  upon  the  scale,  the  lengths  of  the  lines  in  the 
force  diagram  may  be  computed  arithmetically.  The  sizes 
measured  by  the  scale,  when  the  diagram  is  carefully  drawn, 
are  sufficiently  accurate  for  all  practical  purposes ;  but  in 
some  cases,  such,  for  instance,  as  when  the  implements  for 
making  a  correct  diagram  are  not  at  hand,  and  in  all  cases 
as  a  check  upon  the  accuracy  of  the  results  obtained  by  the 
graphic  method,  to  be  able  to  arrive  at  the  correct  results 
arithmetically  would  be  useful.  Preparatorv  to  computing 
r  the  lengths  of  the  lines,  it  will  be  observed  that  the  triangle 
KAJ,  Fig.  127,  is  precisely  proportionate  to  the  triangles 
formed  by  the  inclination  of  the  rafters  of  Fig.  126  with  the 
vertical  and  horizontal  lines ;  that  all  the  inclined  lines  of 
Fig.  127  are  drawn  at  equal  angles  of  elevation  ;  and  that  the 
triangles  formed  by  these  inclined  lines  with  the  vertical 
and  horizontal  lines  are  all  homologous. 

Since  the  height  of  the  roof  is  given  at  20  feet,  and  half 
the  span  is  30  feet,  therefore  the  perpendicular  and  base 
lines  of  each  triangle  are  in  like  proportion — namely,  as  20 
to  30,  or  as  i  to  ij. 

The  perpendicular  being  the  weight  in  each  case,  which 
is  known,  we  may,  therefore,  by  this  proportion  obtain  the 
base.  Having  both  base  and  perpendicular,  the  length  of 
the  hypothenuse  may  be  found  by  Euclid's  47th  of  ist  book — 


COMPUTING   THE    STRAINS.  487 

the  length  of  the  hypothenuse  equals  the  square  root  of  the 
sum  of  the  squares  of  the  base  and  perpendicular.  If  the 
hypothenuse  of  one  triangle  be  computed  by  this  method, 
that  of  the  others  (since  the  triangles  are  homologous)  may 
be  found  by  the  more  simple  method  of  proportion. 

Taking  a  triangle  having  the  perpendicular  and  base 
equal  to  i  and  i|,  we  find,  by  the  above  rule,  that  its 
hypothenuse  equals  1-802776  nearly.  The  hypothenuses  of 
the  other  triangles,  therefore,  may  be  found  by  the  proportion : 

I   :   1-802776  : :  /  :  h 

h  =  i- 802 776^ 
and  for  the  base  we  have 

I   :    i  -  5   : :  /  :  b 
b=  i  -  5/ 

With  these  formulas,  the  lines  in  Fig.  127  have  been  computed. 
The  strains  in  the  proposed  truss  (Fig.  126),  by  both  methods, 
have  been  found  to  be  as  follows  : 

HY    SCALE.  BY    COMPUTATION. 

AK  —  47,800     pounds ;  47,864     pounds. 

BL  =  42,000  42,005 

CM  =  5,800  "  5,859 

AB  =  5,800  "  5,859 

CD  —  30,100  "  30W5 

DH  =  35,000  "  34,950 

AJ  =  39,800  "                '   39,825 

BD  =  13,500  "  13,550 

682. — I>imeiision§  of  Part§  Subject  to  Tension. — With 
these  forces,  and  the  appropriate  rules  hereinbefore  given, 
the  dimensions  of  the  several  parts  of  the  truss  may  now  be 
determined. 


ROOF   TRUSSES.  CHAP.    XXIII. 

Commencing  with  the  tie-beam,  KP,  it  may  be  observed, 
preparatory  to  computing-  its  dimensions,  that  while  this 
piece,  in  resisting  the  thrust  of  the  rafters,  is  subjected  to  a 
tensile  strain,  it  is  also  subject  to  a  transverse  strain  from 
the  weight  of  the  ceiling  and  floor  which  it  has  to  carry. 
These  two  strains,  however,  are  of  such  a  nature  that  in 
their  effect  upon  the  beam  they  do  not  conflict  ;  for  the 
•tensile  strain  from  the  thrust  of  the  rafters,  acting,  as  it  will 
usually,  in  the  upper  half  of  the  beam,  serves  to  counteract 
the  compression  produced  by  the  transverse  strain  in  this 
part  of  the  beam,  and  the  fibres  near  the  middle  of  the 
beam,  owing  to  their  proximity  to  the  neutral  line,  being 
strained  very  little  by  the  transverse  strain,  have  a  large 
reserve  of  strength  available  to  assist  in  resisting  the  tensile 
strain.  It  will  be  sufficient,  therefore,  to  provide  a  piece  of 
timber  for  the  tic-beam  of  sufficient  size  to  resist  only  one 
of  the  two  strains ;  not  necessarily  that  strain,  however, 
which  is  the  greater,  but  that  one  which  requires  the  larger 
piece  of  timber  to  resist  it. 

The  computations  of  dimensions  required  to  resist  the 
two  strains  will  now  claim  attention. 

For  the  tensile  strain  we  have,  'by  formula  (299.)^ 

20  x  39800  _ 
16000" 

or  say    50   inches  area  of  cross-section,    for  Georgia    pine. 
For  white  pine  the  area  should  be    65     inches, 

The  load  producing  transverse  strain  is  (Art.  678)  10,300 
pounds.  The  rule  for  determining  the  proper  area  of  cross- 
section  is  to  be  found  in  formula  (130.\  which  may  be 
modified  for  this  case  by  substituting  rl  for  <5,  the  symbol 
for  deflection,  and  by  putting  for  r  the  rate  0-04  of  an 
inch.  With  these  substitutions,  we  have 


STRAINS   IN   TIE-BEAM.  489 

Fixing  upon  a  proportion  for    b    in  terms  of  d,    say,  for  ex- 
ample,   b  —  f^>    and  substituting  this  value  for    b,    we  have 

IU12  —  0-04  x  IFd* 
MfWV     ,< 

F  ~ 

If    the  timber  is  to  be  of  white    pine,    then     F    equals 
2900     (Table  XX.),  and  we  have 


4/2O| 

=r  y  - 


X  10300  X  20  . 

-  =  13-  116 


2900 

or  the  depth  will  need  to  be  I3-J-  inches.  Three  quarters 
of  this,  or  9!,  will  be  the  breadth.  The  tie-beam,  of  white 
pine,  will  need  to  be,  therefore,  say  lox  13  inches.  If  of 
Georgia  pine,  instead  of  white  pine,  then  5900,  the  value  of 
F  for  Georgia  pine,  must  be  substituted  for  2900  in  the 
formula,  and  the  results,  8-237  and  10-982,  will  show,  say 
8J-  x  1  1  inches  as  the  size  of  timber  required. 

The  dimensions  thus  found,  to  resist  the  transverse  strain, 
being  in  excess  of  those  required  to  resist  the  tensile  strain, 
are  to  be  adopted  as  the  dimensions  of  the  required  tie- 
beam. 

The  length  of  the  tie-beam,  60  feet,  being  greater  than 
can  readily  be  obtained  in  one  piece,  it  will  have  to  be  built 
up.  In  doing  this,  it  is  necessary  that  each  piece  be  of  the 
full  height  of  the  beam,  or  that  the  joints  of  the  make-up  be 
vertical  and  not  horizontal.  These  vertical  laminas  should 
be  in  pieces  of  such  lengths  that  no  two  heading  joints  occur 
within  five  feet  of  each  other,  and  that  these  joints  shall  be 
as  near  as  practicable  to  the  two  vertical  suspending  rods. 
The  laminas  need  to  be  well  secured  together  with  proper 
iron  bolts.  The  feet  of  the  rafters  should  be  provided  with 
iron  clamps  of  sufficient  area  to  resist  the  horizontal  strain 
there,  and  should  be  secured  to  the  tie-beam  with  bolts  of 
corresponding  resistance. 


490  ROOF   TRUSSES.  CHAP.    XXIII. 

If  the  iron  in  the  bolts  and  clamps  of  the  truss  be  of  aver- 
age good  quality,  it  may  be  calculated  on  as  resisting  effec- 
tually 9000  pounds  per  square  inch  (see  Art.  642).  The 
vertical  suspension  rods  BD  and  DE,  Fig.  126,  may  also 
be  calculated  for  a  like  strain. 


683. — Dimen§ion§  of  Part§  Subject    to    €ompre§§ion. — 

The  rafters,  straining  beam  and  braces  are  all  subject  to 
compression,  and  their  dimensions  may  now  be  obtained. 

The  areas  of  these  pieces  may  be  had  by  the  use  of 
formula  (301.)  ;  or,  as  this  in  some  cases  is  objectionable,  for 
the  reason  that  the  ratio  between  the  length  and  thickness 
has  to  be  assumed  in  advance,  we  may  find  in  formula  (303.) 
a  rule  free  from  this  objection,  but  encumbered  with  more 
intricate  computations.  Formula  (301.),  when  used  by  those 
having  experience  in  such  work,  is  far  preferable^  on  account 
of  its  greater  simplicity. 

Taking  first  the  rafter,  and  the  portion  of  it  at  the  foot, 
where  the  strain  is  greatest,  47,800  pounds,  we  have  for  its 
length  about  12  feet.  If  of  Georgia  pine,  its  thinnest 
dimension  of  cross-section  will  probably  be  about  8  inches. 

Then     r=  -  =  18     (see  Art.  643).     The   value 

of  C  is  9,500  and  the  value  of  e  is  0-00109,  both  by 
Table  XX.  Making  the  symbol  for  safety,  a,  equal  10 
we  have 


io[i  +(f  x  0-00109  x  1 82)]  47800 

A  — — —    — —  m  70  • 

9500  / 


or  the  area  of  the  rafter  should  be    77,    say     8  x  9f     inches. 
If  computed  by  formula  (303.),  putting     n  =  1-2,     the 
exact  size   will  be  found  at     8-006  x  9-607  =  76-92     inches 
area. 


PIECES   SUBJECT   TO    COMPRESSION. 

684.  _  I>imeii§ion§  of  Mid-Rafter.  —  In  the  rafter  at  BL 
the  strain  is  42,000  pounds.  The  length  and  ratio  here  will 
be  the  same  as  at  AK,  and  the  dimensions  of  AK  and 
BL  are  therefore  in  proportion  to  the  weights  (form. 
301.\  or 

47800  :  42000  :  :  76-92  :  A 


so  that    68    inches  of  sectional  area,  or    8  x  8J    inches,  is  the 
size  required. 


685. — Dimension  of  Upper  Rafter. — The  upper  end  of 
the  rafter  has  only  the  weight  at  the  ridge,  5,800  pounds,  to 
bear.  The  thickness  of  the  rafter  here  will  probably  be  but 
4  inches.  This  gives  a  ratio  of  i|^  =  36.  With  this  ratio, 
with  5,800  for  the  weight,  and  with  the  other  quantities  as 
before,  a  computation  by  formula  (301^  will  result  in  show- 
ing the  required  area  to  be  19-04,  or,  say  4x5  inches; 
but,  in  order  to  resist  effectually  the  distributed  load  of  the 
roofing,  this  part  of  the  rafter  should  not  be  less  than  4x8 
inches. 


686. — l>imen§ion§  of  Braee. — The  brace,  AB,  being  of 
equal  length  and  carrying  an  equal  load  with  the  upper  end 
of  the  rafter,  may  be  made  of  the  size  there  found  necessary, 
or,  say  4x6  inches. 


687. — Dimeiiiions    of    Straining-Beam. — The    straining- 
beam     CD     is  compressed  with  a  strain  of    30,100     pounds, 


492  ROOF  TRUSSES.  CHAP.   XXIII. 

and  its  length  is     20    feet. 

Assuming  its  thickness  to  be  that  of  the  rafter,  we  have 

r  = —  =  30,     and  in  formula  (301.) 

)1 30100 

-=78-29 
9500 

or  its  area  should  be     ;8i,     or,  say     8  x  10=  80     inches. 

With  this  result,  the  computation  of  the  dimensions  of  all 
the  pieces  of  the  truss  is  completed  ;  for  the  other  rafter  and 
brace  are  in  like  condition  with  those  computed,  and  should 
therefore  be  of  the  same  dimensions. 


QUESTIONS   FOR   PRACTICE. 


688. — In  a  roof  truss  similar  to  that  shown  in  Fig.  109,  of 
42  feet  span  and  14  feet  height,  measuring  from  the 
axial  lines :  What  will  be  the  strains  in  the  various  pieces  of 
the  truss,  with  a  load  of  5,000  pounds  at  each  of  the  three 
points  above  the  rafters,  and  a  load  of  10,000  pounds 
suspended  from  the  centre  of  the  tie-beam  ? 

Draw  the  appropriate  force  diagram,  and  give  the  strains 
from  measurement. 


689. — Draw  a  force  diagram  for  a  roof  truss  similar  to 
the  design  in  Fig.  in,  with  a  span  of     54     feet  and  a  height 


QUESTIONS   FOR   PRACTICE.  493 

of  1 8  feet;  the  upper  weights  being  taken  at  6,000 
pounds  each,  the  central  weight  under  the  tie-beam  at 
5,000  pounds,  and  each  of  the  two  other  weights  at  7,000 
pounds. 

Show,  from   the  diagram,  the  strain  in  each  line  of  the 
truss. 


690. — In  a  truss  similar  to  that  in  Fig.  121,  show,  by  a 
force  diagram,  what  would  be  the  strains  in  each  line,  when 
the  span  is  40  feet  and  the  height  20  feet.  The  weights 
FG  and  GH  are  so  located  as  to  divide  the  span  into 
three  equal  parts,  the  three  loads  above  the  rafters  are  each 
7,000  pounds,  and  the  two  loads  below  each  4,000  pounds. 
The  point  JABK  is  to  be  taken  at  the  middle  of  the 
rafter,  and  the  line  AB  is  to  be  drawn  at  right  angles 
with  the  rafter. 

691. — In  a  roof  with  an  elevated  tie-beam,  such  as  in  Fig. 
125,  with  a  span  of  40  feet  and  height  of  20  feet,  and 
with  the  tie  elevated  at  the  middle  8  feet  above  the  level 
of  the  feet  of  the  rafters,  compute  the  strain  in  the  suspen- 
sion-rod at  the  middle,  due  to  the  elevation  of  the  tie  ;  the 
weight  upon  one  half  of  the  truss  being  24,000  pounds. 

692. — In  a  building  119  feet  long,  and  80  feet  wide 
to  the  centres  of  bearings,  and  having  the  side  walls  pierced 
for  seven  windows  each,  state  how  many  roof  trusses  there 
should  be. 

Which  of  the  designs  given,  having  a  tie  horizontal  from 
the  feet  of  the  rafters,  would  be  appropriate  for  the  case  ? 

The  roof  is  to  be  25  feet  high  at  middle,  and  to  have 
the  interior  space  along  the  middle  free  from  timber.  The 


494  ROOF   TRUSSES.  CHAP.    XXIII. 

load  upon  the  roof  is  to  be  taken  at  50  pounds  per  foot 
horizontal,  upon  the  tie-beam  at  40  pounds  to  the  foot, 
and  upon  the  straining  beam  at  5  pounds  per  foot. 

Make  a  force  diagram,  and   from   it  show  the  strains  in 
each  piece. 

.Compute  the  dimensions  of  the  several  timbers,  which 
are  all  to  be  of  Georgia  pine  ;  the  rafter  being  9  inches 
thick  below  the  straining-beam  and  6  inches  above,  and 
the  iron  work  being  subjected  to  a  tensile  strain  of  9000 
pounds  per  inch. 


CHAPTER    XXIV. 

TABLES. 

ART.  693.— Tables  I.  to  XXI.— Their  Utility.— Rules  for 

determining  the  required  dimensions  of  the  various  timbers 
in  floors  are  included  in  previous  chapters.  These  rules  are 
carefully  reduced  to  the  forms  required  in  practice.  In  using 
them,  it  is  only  needed  to  substitute  for  the  various  alge- 
braic symbols  their  proper  numerical  values,  and  to  perform 
the  arithmetical  processes  indicated,  in  order  to  arrive  at  the 
result  desired. 

To  do  even  this  simple  work,  however,  requires  care  and 
patience,  and  these  the  architect,  owing  to  the  multiplicity 
of  detail  demanding  time  and  attention  in  his  professional 
practice,  frequently  finds  it  difficult  to  exercise.  To  relieve 
him  of  this  work,  the  first  twenty-one  of  the  following  tables 
have  been  carefully  computed.  Tables  I.  to  XXI.  afford  the 
data  for  ascertaining  readily  the  dimensions  of  the  beams 
and  principal  timbers  required  in  floors  of  dwellings  and  first- 
class  stores.  Tables  XVII.,  XVIII.  and  XIX.  refer  to 
beams  of  rolled-iron  ;  the  others  to  those  of  wood. 

694. — Floor  Beams  of  Wood   and  Iron  (I.  to  XIX.  and 

XXI.) — In  these  tables  will  be  found  the  dimensions  of  Floor 
Beams  and  Headers,  of  Hemlock,  White  pine,  Spruce  and 
Georgia  pine  ;  for  Dwellings  and  for  First-class  stores. 

Tables  XVIII.  and  XIX.  exhibit  the  distances  from  cen- 
tres at  which  Rolled-iron  Beams  are  required  to  be  placed 


496  TABLES.  CHAP.  XXIV. 

in    Banks,    Office   Buildings   and  Assembly-Rooms,  and  in 
First-class  Stores. 


695. — Floor  Beams  of  Wood  (I.  to  Till.). — In  these 
tables  the  recorded  distance  from  centres  is  in  inches,  and  is 
for  a  beam  one  inch  thick,  or  broad.  The  required  distance 
from  centres  is  to  be  obtained  by  multiplying  the  tabular  dis- 
tance by  the  breadth  of  the  given  beam. 

For  example  :  Let  it  be  required  to  ascertain  the  dis- 
tance from  centres  at  which  white  pine  3  x  10  inch  beams, 
1 6  feet  long  in  the  clear  of  the  bearings,  should  be  placed 
in  a  dwelling. 

By  reference  to  Table  II.,  "  White  Pine  Floor  Beams  One 
Inch  Thick,  for  Dwellings,  Office  Buildings,  and  Halls  of 
Assembly,"  we  find,  vertically  under  10,  the  depth,  and 
opposite  to  16,  the  length,  the  dimension  4-5.  This  is 
the  distance  from  centres  for  a  beam  one  inch  broad.  Then, 
since  the  given  beam  has  a  breadth  of  3  inches, 

3><4-5=  13-5 

equals  the  required  distance  from  centres  for  beams  3 
inches  broad.  Therefore,  3  x  10  inch  white  pine  beams 
with  16  feet  clear  bearing,  should,  in  a  dwelling,  etc.,  be 
placed  13^  inches  from  centres. 

Tables  I.  to  IV.  were  computed  from  formula  (14$-), 

cl*  =  ibd3 

which,  with     b=  I,     and  putting     c     in  inches,  becomes 

'  (308.) 


FLOOR   BEAMS   AND    HEADERS.  497 

Tables  V.  to  VIII.  were  computed  from  formula  (149.), 

cl3  =  kbds 
which,  with     £=i,     and  with     c    in  inches,  becomes 

\2kd3 


c  =  - 


(307.) 


696.—  Hcader§  of  Wood  (IX.  to  XVI.).—  (See  Art.  142.) 
The  results  recorded  in  these  tables  show  the  breadth  of 
headers  which  carry  tail  beams  one  foot  long.  The  tabular 
breadth,  if  multiplied  by  the  length  in  feet  of  the  given  tail 
beam,  will  give  the  breadth  of  the  required  header. 

For  example  :  Let  it  be  required  to  ascertain  the  breadth 
of  a  Georgia  pine  header  20  feet  long,  15  inches  deep, 
and  carrying  tail  beams  12  feet  long,  in  the  floor  of  a  first- 
class  store.  By  referring  to  Table  XVI.,  4'  Georgia  Pine 
Headers  for  First-class  Stores,"  at  the  intersection  of  the 
vertical  column  for  15  inches  depth  and  the  horizontal 
line  for  20  feet  length,  we  find  the  dimension  i«o6.  This 
is  the  breadth  of  the  header  for  each  foot  in  length  of  the 
tail  beams.  As  the  tail  beams  in  this  case  are  12  feet  long, 
therefore  12x1-06=12.72,  equals  the  required  breadth  of 
the  header  in  inches. 

The  first  four  (IX.  to  XII.)  of  these  tables  were  computed 
from  formula  (156.), 


i6Fr(d-i? 

which,    when    reduced     (putting       r  =  0-03,      /  =  90       and 
n  =  i)     becomes 


498  TABLES.  CHAP.    XXIV. 

The  second  four  (XI  H.  to  XVI.)  of  these  tables  were  com- 
puted from  the  same  formula,  (156-},  by  putting  r  =  0-04, 
f=  275  and  n  =  i  ;  which  reduction  gives 


(309, 


697. — Elements  of  Rolled-Iron  Beams  (XVII.). — Table 
XVII.  contains  the  dimensions  of  cross-section  and  the 
values  of  /,  the  moment  of  inertia,  for  1 19  of  the  rolled- 
iron  beams  of  American  manufacture  in  use.  These  values 
are  required  in  using  the  rules  in  Chapter  XIX.,  by  which 
the  capacities  of  the  beams  are  ascertained.  (See  Arts.  479 
to  482,  485  to  492,  501,  511,  512,  514,  517,  519,  521,  523, 
etc.) 

The  values  of    /    were  computed  by  formula  (213.) 


698.  —  Rolled-Iron     Beams     for     Office     Buildings,     etc. 

(XVIII.).  —  Table  XVI  II.  contains  the  distances  from  centres, 
in  feet,  at  which  rolled-iron  beams  should  be  placed,  in  the 
floors  of  Dwellings,  Banks,  Office  Buildings  and  Assembly 
Halls.  (See  Arts.  500  and  501.) 

These  distances  were  computed  by  formula  (237.), 


__    y_ 
'  I3  420 


699. — Rolled-Iron  Beams  for  First-class  Stores  (XIX.).— 

Table  XIX.  contains  the  distances  from  centres,  in  feet,  at 
which  rolled-iron  beams  should  be  placed,  in  the  floors  of 
First-class  Stores.  (See  Arts.  504  and  505.) 


ROLLED-IRON   BEAMS.  499 

These  distances  were  computed  by  formula  (239.\ 
148-87        y 


I3  960 


700. — Example. — As  an  example  to  show  the  uses  of 
Tables  XVIII.  and  XIX. :  Let  it  be  required  to  know  the  dis- 
tances from  centres  at  which  9  inch  84  pound  Phcenix 
rolled-iron  beams  should  be  placed,  on  walls  with  a  span  or 
clear  bearing  of  18  feet,  to  form  a  floor  to  be  used  in  an 
Office  Building  or  Assembly  Room. 

In  Table  XVIII.,  the  one  suitable  for  this  case,  at  the 
intersection  of  the  vertical  column  for  18  feet,  with  the 
horizontal  line  for  the  given  beam  named  above,  we  find 
4-51,  or  4^  feet,  the  required  distance  from  centres. 

For  a  First-class  Store  (see  Table  XIX.),  these  beams,  if 
of  the  length  stated,  should  be  placed  2-66,  or  2  feet 
and  8  inches  from  centres. 


701. — Constants  for  Use  in  the  Rules  (XX.).— Constants 
for  use  in  the  rules  in  previous  chapters  are  to  be  found  in 
Table  XX. 

These  constants,  for  the  13  American  woods  named 
and  for  mahogany,  have  been  computed  from  experiments 
made  by  the  author  in  1874  and  1876  expressly  for  this  work 
(Arts.  704  to  707).  For  the  values  of  B  and  F,  the 
lowest  and  highest  of  the  two  series  of  experiments  are 
taken,  and  the  average  given  for  use  in  the  rules. 

The  constants  for  the  other  woods  named  in  the  table 
have  been  computed  for  this  work  from  experiments  made 
by  Barlow,  and  recorded  in  his  work  on  the  Strength  of 
Materials. 

The  constant     F,     for  American  wrought-iron,  was  com- 


5OO  TABLES.  CHAP.  XXIV. 

puted  by  the  author  from  six  tests  made  by  Major  Anderson 
on  rolled-iron  beams  at  the  Trenton  Iron  Works,  and  from 
two  tests  made  at  the  works  of  the  Phoenix  Iron  Co.  of 
Philadelphia.  The  beams  upon  which  these  tests  were  made 
were  from  6  to  15  inches  deep  and  from  12  to  27 
feet  long. 

The  values  of  F  for  the  other  metals,  ajid  of  B  for 
all  the  metals,  have  been  computed  from  tests  made  by  trust- 
worthy experimenters,  such  as  Hodgkinson,  Fairbairn,  Kir- 
kaldy,  Major  Wade  and  others.  The  average  of  these 
values  may  be  used  in  the  rules,  for  good  ordinary  metal. 
For  any  important  work,  however,  constants  should  be 
derived  from  tests  expressly  made  for  the  work,  upon  fair 
specimens  of  the  particular  kind  of  metal  proposed  to  be 
used. 


702.— Solid  Timber  Floors  (XXI.).— The  depths  re- 
quired for  beams  when  placed  close  to  each  other,  side  by 
side,  without  spaces  between  them,  may  be  found  in  Table 
XXI. 

This  is  not  an  economical  method  of  construction.  More 
timber  is  required  than  in  the  ordinary  plan  of  narrow,  deep 
beams,  set  apart.  But  a  solid  floor  has  the  important 
characteristic  of  resisting  the  action  of  fire  nearly  as  long, 
if  not  quite,  as  a  floor  made  with  rolled-iron  beams  and 
brick  arches. 

A  floor  of  timber  as  usually  made,  with  spaces  between 
the  beams,  resists  a  conflagration  but  a  very  short  time. 
The  beams  laid  up  like  kindling-wood,  with  spaces  between, 
afford  little  resistance  to  the  flames  ;  but,  when  laid  close, 
they,  by  the  solidity  obtained,  prevent  the  passage  of  the 
air.  The  fire,  thus  retarded  and  confined  to  the  room  in 
which  it  originated,  may  be  there  extinguished  before  doing 


SOLID   TIMBER   P^LOORS.  5<DI 

serious  damage.  Floors  built  solid  should  be  plastered 
upon  the  underside.  The  plastering  lath  should  be  nailed  to 
narrow  furring  strips,  half  an  inch  thick,  and  the  plastering 
pressed  between  the  lath  so  as  to  till  the  half  inch  space 
with  mortar.  The  mortar  used  should  contain  a  large 
portion  of  plaster  of  Paris,  and  be  finished  smooth  with  it. 
Owing  to  the  fire-proof  quality  of  this  material,  it  will  pro- 
tect the  lath  a  long  time.  Thus  constructed,  a  solid  floor 
will  possess  great  endurance  in  resisting  a  conflagration. 

The  timbers  should  be  attached  to  each  other  by  dowels. 
These  will  serve,  like  cross-bridging,  to  distribute  the 
pressure  from  a  concentrated  weight  to  the  contiguous 
beams. 

The  depths  given  in  Table  XXI.  were  computed  by 
formulas  (311.)  and  (312.).  These  were  reduced  from  form- 
ula (130.).  which  is 


In  this  formula     U  —  cfl,     c    and     /    being   taken  in   feet. 
If    c    be  taken  in  inches,   then   for     c     we  have     —  ,     and 

U=-fL     Putting    rl    for    8    (Art.  313)  we  have 


In  a  solid  floor  the  breadth  of  the  beams  will  equal  the 
distances  from  centres,  or  b  —  c  (c  now  being  in  inches). 
In  the  formula  these  cancel  each  other  ;  or 

?   =  Fd'r  and 


8x  12 

<*3=~l;  («*) 


502  TABLES.  CHAP.  XXIV. 

For  dwellings  and  halls  of  assembly,  we  have  taken 
(Art.  115)  f  at  90,  or  70  for  the  superincumbent  load 
and  20  for  the  materials  of  construction.  In  a  solid  floor, 
however,  the  weight  of  the  timbers  differs  too  much  to 
permit  an  average  of  it  to  be  used  as  a  constant  in  the 
formula.  The  weight  of  the  plastering,  furring  and  floor- 
plank  is  constant,  and  may  be  taken  at  12  pounds.  To 
this  add  70  for  the  superincumbent  load,  and  the  sum,  82, 
.plus  the  weight  of  the  beam,  will  equal  /,  the  total  load. 

The  weight  of  the  beam  will  equal  the  weight  of  a  foot 
superficial,  inch  thick,  of  the  timber,  multiplied  by  the  depth 
of  the  beam  ;  or,  putting  y  equal  to  the  weight  of  one 
foot,  inch  thick,  of  the  timber,  we  have  its  total  weight  equal 
to  yd  ;  or,  •  f  =  %2  +  yd.  Substituting  this  value  for  f  in 
formula  (310.),  and  putting  r  =  0-03,  then  we  have 


19-2  x  o-o^F 


This  formula  is  general  for  floors  of  dwellings,  office 
buildings,  and  halls  of  assembly.  As  the  symbol  for  the 
depth  is  found  on  both  sides  of  this  equation,  the  depth  for 
any  given  length  can  not  be  directly  obtained  by  it  ;  a  modi- 
fication is  needed  to  make  the  formula  practicable. 

An  inspection  of  the  formula  shows  that  the  depth  will 
be  very  nearly  in  direct  proportion  to  the  length.  By  a 
simple  transformation  of  the  symbols,  a  formula  is  obtained 
which  will  give  the  length  for  any  given  depth.  By  an  ap- 
plication of  this  formula  to  the  two  extremes  of  depth  and 
length  for  each  kind  of  material,  the  relative  values  of  d 
and  /  may  be  found.  The  results  for  the  two  extremes  in 
each  case  will  differ  but  little.  An  average  may  be  used  as 
a  constant  for  all  practical  lengths,  without  appreciable 


THICKNESS    OF   SOLID    TIMBER   FLOORS.  503 

error.     The  values  of    d    have  been  computed  for  the  four 
woods  named  below,  and  the  average  value  found  to  be  for 

Georgia  pine,  d  =  O-3I4/ 

Spruce,  d  =.  o-$6$l 

White  pine,  d  =  0-3897 

Hemlock,  d  —  o-^gl 

An  average  value  of  y,  the  weight  per  foot  superficial, 
inch  thick,  may  be  taken  as  follows  :  for 

Georgia  pine,  y  —  4 

Spruce,  y  =  2% 

White  pine.  y  =  2-J 

Hemlock,  y  —  2 

With  these  values  of  y  and  d,  formula  (311.}  becomes 
practicable,  and  will  give  the  required  depth  for  any  given 
length  of  floor  beams,  of  the  four  woods  named,  for  the 
solid  floors  of  dwellings,  office  buildings,  and  halls  of 
assembly. 

For  the  floors  of  first-class  stores,  taking  250  pounds 
as  the  superincumbent  load  and  13  pounds  as  the  weight 
of  the  plastering,  flooring,  etc.,  and  putting  r  =  0-04  we 
have,  in  formula  (310.), 


This  formula  is  general  for  floors  of  first-class  stores.  The 
values  of  d  have  been  computed  for  the  extremes  of 
lengths,  and  an  average  found  to  be  as  follows  :  for 

Georgia  pine,  d  =  -4/ 

Spruce,  d  = 

White  pine,  d  = 

Hemlock,  d  =  -so6/ 


504  TABLES.  CHAP.  XXIV. 

With  these  values  of  d,  and  the  above  values  of  y,  for- 
mula (312.)  will  give  the  depths  of  solid  floors  for  first-class 
stores. 

The  depths  of  solid  floors  in  Table  XXL,  for  dwellings, 
office-buildings  and  halls  of  assembly,  were  computed  by 
formula  (311.),  and  those  for  first-class  stores  by  formula 
(312.) 

703.— Weights  of  Building  Materials  (XXII.).— Table 
XXII.  contains  the  weight  per  cubic  foot  of  various  build- 
ing materials. 

704. — Experiments     on    American     Woods    (XXIII.     to 

XE,  VI.).— Tables  XXIII.  to  XLVL,  inclusive,  contain  the 
results  of  experiments  upon  six  of  our  American  woods  such 
as  are  more  commonly  used  as  building  material. 

These  experiments,  as  well  as  those  of  1874  (Art.  701), 
were  made  upon  a  testing  machine  constructed  for  the 
author,  and  after  his  plan,  by  the  Fairbanks  Scale  Co.  It  is 
a  modification  of  the  Fairbanks  scale,  a  system  of  levers 
working  on  knife  edges,  and  arranged  with  gearing  and 
frame  by  which  a  very  gradual  pressure  is  brought  to  bear 
upon  the  piece  tested,  which  pressure  is  sustained  by  the 
platform  of  the  scale  and  thus  measured. 

By  an  application  of  clock-work,  devised  by  Mr.  R.  F. 
Hatfield,  son  of  the  author,  the  poise  upon  the  scale  beam  is 
kept  in  motion  by  the  pressure  upon  the  platform,  and  is 
arrested  at  the  instant  of  rupture  of  the  piece  tested.  For 
the  moderate  pressures  (under  2000  pounds)  required, 
this  machine  is  found  to  work  satisfactorily. 

705. — Experiments  by  Transverse  Strain  (XXIII.  to 
XXXV.,  Xi.il.  and  XLIII.).— Tables  XXIII.  to  XXXV. 


EXPERIMENTS   ON   WOODS.  505 

contain  tests  by  Transverse  Strain,  upon  six  of  the  thirteen 
woods  tested  by  the  author  for  this  work. 

At  intervals,  as  shown,  the  pressure  was  removed  and  the 
set,  if  any,  measured.  It  was  found  that  in  many  instances 
a  decided  set  had  occurred  before  the  increments  of  deflec- 
tion had  ceased  being  equal  for  equal  additions  of  weight. 
It  was  thus  made  plain  that  some  modification  of  this  rule 
for  determining  the  limit  of  elasticity  must  be  made.  To 
fix  this  limit  clearly  inside  of  any  doubtful  line,  25  per 
cent  of  the  deflection  obtained,  while  the  increments  of  de- 
flection remained  equal  for  equal  additions  of  weight,  was 
deducted,  and  the  remainder  taken  as  the  deflection  at  the 
limit  of  elasticity. 

With  this  deflection,  the  values  of  the  constants  e  and 
a  in  Table  XX.  were  computed  (Art.  701). 

The  load  upon  a  beam,  determined  by  the  rules  with  the 
constants  restricted  within  this  limit,  will  not,  it  is  confident- 
ly believed,  be  subject  to  set ;  or  if,  as  is  claimed  by 
Professor  Hodgkinson,  any  deflection,  however  small,  will 
produce  a  set,  that  this  set  will  be  so  slight  and  of  such  a 
nature  as  not  to  be  injurious,  or  worthy  of  consideration. 

A  resume  of  the  results  of  Tables  XXIII.  to  XXXV.  is 
given  in  Tables  XLII.  and  XLIII. 

The  values  of  F  and  B,  given  in  Table  XX.,  were 
derived,  not  alone  from  the  results  given  in  these  tables, 
but  also  from  results  of  the  other  experiments  made  in  1874. 
(Art.  701.) 


706.  —  Experiments  by  Tensile  and  Sliding  Strains 
(XXXVI.  to  XXXIX.,  XL.IV.  and  XLV.).— Tables  XXXVI.  and 
XXXVII.  contain  tests  of  the  resistance  to  tensile  strain  of 
six  of  the  more  common  American  woods. 

A  rhiimt  of  the  results  is  given  in  Table  XLI V. 


506  TABLES.  CHAP.  XXIV. 

Tables  XXXVIII.  and  XXXIX.  give  tests  made  to  show 
the  resistance  to  sliding  of  the  fibres  in  six  of  the  more 
common  American  woods.  These  experiments  were  made 
to  ascertain  the  power  of  the  several  woods  to  resist  a  force 
tending  to  separate  the  fibres  by  sliding,  in  the  longitudinal 
direction  of  the  fibres.  The  rafter  of  a  roof,  when  stepped 
into  an  indent  in  the  tie-beam,  exerts  a  thrust  tending  to 
split  off  the  upper  part  of  the  end  of  the  tie-beam.  A  pin 
through  a  tenon,  when  subjected  to  strain,  tends  to  split  out 
the  part  of  the  tenon  in  front  of  it.  These  are  instances  in 
which  rupture  may  occur  by  the  sliding  of  the  fibres  longi- 
tudinally, and  a  knowledge  of  the  power  of  the  various 
woods  to  resist  it,  as  shown  in  these  tables,  and  as  condensed 
in  Table  XLV.,  will  be  useful  in  apportioning  parts  subject 
to  this  strain.  The  symbol  G,  in  Table  XX.,  represents 
in  pounds  the  sliding  resistance  to  rupture  per  square  inch 
superficial,  and  is  equal  to  the  average  of  the  results  of  the 
experiments  in  Table  XLV.  A  discussion  to  show  the 
application  of  these  results  is  omitted  as  being  uncalled  for 
in  a  work  on  the  Transverse  Strain.  For  its  treatment,  see 
"  American  House  Carpenter/'  Arts.  301  to  303,  where  H, 
the  value  of  each  wood,  is  taken  at  \  of  the  resistance  to 
rupture. 

707. — Experiments  by   Crushing   Strain    (XL.,  XLI.  and 

XL.VI.). — Tables  XL.  and  XLI.  contain  tests  of  resistance  to 
crushing,  in  the  direction  of  the  fibres,  of  six  of  the  more 
common  of  our  American  woods.  The  pieces  submitted  to 
this  test  were  from  one  to  two  diameters  high. 

A  rtsumt  of  the  results  is  given  in  Table  XLVL 


TAB  L  E  S. 


TABLE    I. 


HEMLOCK   FLOOR   BEAMS    ONE    INCH   THICK,    FOR   DWELLINGS, 
OFFICE  BUILDINGS,   AND    HALLS   OF   ASSEMBLY. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams  Thicker  than  One  Inch,  see  Arts.  693  and  695. 


LENGTH 

DEPTH  OF  BEAM  (in  inches). 

BETWEEN 

BEARINGS 

(in  feet). 

6 

7 

8 

9 

1O 

11 

12 

13 

14 

7 

n-3 

8 

7-6 

12-0 

9 

5-3 

8-4 

12-6 

10 

3-9 

6-1 

9-2 

I3-I 

11 

2-9 

4-6 

6-9 

9-8 

12 

.. 

3-6 

5-3 

7-6 

10-4 

13 

.. 

2-8 

4-2 

5-9 

8-2 

10-9 

14 

•  ••. 

3-3 

4-8 

6-5 

8-7 

n-3 

15 

,•• 

2-7 

3-9 

5-3 

7-i 

9-2 

u-7 

16 

3-2 

4-4 

5-8 

7-6 

9-6 

17 

.  , 

.. 

2-7 

3-6 

4-9 

6-3 

8-0 

IO-O 

18 

•• 

.. 

.. 

3-r 

4-1 

5-3 

6-7 

8-4 

19 

.. 

.. 

2-6 

3-5 

4-5 

5-7 

7-2 

20 

•• 

•• 

•• 

•• 

•• 

3-o 

3-9 

4.9 

6-1 

21 

3.  <j 

4.  2 

5.  o 

22 

o 

2-Q 

• 
•5.7 

J 
4-6 

23 

•    V 

O    I 

3.  o 

*T 

4-O 

24 

* 
2-8 

** 

3-6 

508 


TABLE    II. 


WHITE   PINE  FLOOR  BEAMS  ONE  INCH  THICK,  FOR  DWELLINGS, 
OFFICE   BUILDINGS,   AND    HALLS    OF  ASSEMBLY. 

DISTANCE  FROM  CENTRES  (in  inches}. 
For  Beams   Thicker  than  One  Inch,  see  Arts.  693  and  695. 


LENGTH 

DEPTH  OF  BEAM  (in  inches}. 

BETWEEN 

BEARINGS 

(in  feet}. 

6 

7 

8 

9 

10 

11 

12 

13 

14 

7 

n-7 

8 

7-8 

12-4 

9 

5-5 

8-7 

13-0 

10 

4-0 

6-4 

9'5 

11 

3-o 

4-8 

7-i 

10-2 

12 

43-7 

5-5 

7-8 

10-7 

13 

2-9 

4-3 

6-2 

8-4 

II-2 

14 

.. 

3-5 

4-9 

6-8 

9-0 

H.  7 

15 

•• 

2-8 

4-0 

5-5 

7-3 

9'5 

16 

.  . 

3-3 

4-5 

6-0 

7-8 

IO-O 

17 

2-8 

3-8 

5-o 

6-5 

8-3 

10-4 

IS 

.. 

3.2 

4-2 

5-5 

7-0 

8-7 

19 

.. 

2-7 

3-6 

4-7 

5-9 

7.4 

2O 

•• 

•• 

•• 

•• 

3-1 

4-0 

5-i 

6-4 

21 

2-7 

3-5 

4.4 

5-5 

22 

3-0 

3-8 

4-8 

23 

a.  4 

4-2 

24 

2-9 

3-7 

509 


TABLE   III. 


SPRUCE    FLOOR    BEAMS    ONE    INCH    THICK,    FOR    DWELLINGS, 
OFFICE   BUILDINGS,   AND    HALLS   OF  ASSEMBLY. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams  Thicker  than  One  Inch,  see  Arts.  693  and  695. 


LENGTH 

BETWEEN 

BEARINGS 

(in  feet}. 

DEPTH  OF  BEAM  (in  inches}. 

6 

7 

8 

9 

10 

11 

12 

13 

14 

7 

14-1 

8 

9'4 

9 

6-6 

10-5 

1O 

4-8 

7-7 

n-5 

11 

3-6 

5-8 

8-6 

12-3 

12 

2-8 

4-4 

6-6 

9-4 

13 

•  • 

3-5 

5-2 

7'4 

TO  -2 

14 

•  • 

2-8 

4-2 

6-0 

8-2 

10-9 

15 

•• 

•• 

3-4 

4-8 

6-6 

8-8 

n-5 

16 

2-8 

4-0 

5-5 

7-3 

9'4 

17 

.. 

.. 

3-3 

4-6 

6-1 

7-9 

IO-O 

18 

.. 

.. 

.. 

2-8 

3-8 

5-i 

6-6 

8-4 

10-5 

19 

.. 

.. 

.. 

.. 

3-3 

4-3 

5-6 

7-2 

9-0 

2O 

•• 

•• 

•  • 

•'• 

2-8 

3-7  :     4-8 

6-2 

7-7 

21 

3-2  \     4-2 

5-3 

6-6 

22 

.. 

.. 

2-8 

3-6 

4-6 

5-8 

23 

a.'?. 

4.0 

c  .  i 

24 

•'• 

•• 

2-8 

3-6 

4-4 

510 


TABLE   IV. 


GEORGIA   PINE   FLOOR  BEAMS   ONE   INCH   THICK,  FOR  DWELL- 
INGS,  OFFICE   BUILDINGS,   AND   HALLS   OF  ASSEMBLY. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams  Thicker  than  One  Inch,  see  Arts.  693  and  695. 


LENGTH 

BETWEEN 

BEARINGS 

(in  feet). 

DEPTH  OF  BEAM  (in  inches). 

6 

7 

8 

9 

10 

11 

12 

13 

14 

9 

II-2 

10 

8-2 

13-0 

11 

6-1 

9-7 

12 

4'7 

7'5 

II-2 

13 

3-7 

5-9 

8-8 

14 

3-o 

4-7 

7-0 

1O-O 

15 

•• 

3-8 

5-7 

8-2 

II-2 

16 

3-2 

4-7 

6-7 

9-2 

17 

18 

2-6 

3'9 
3-3 

5-6 
4-7 

7-7 
6-5 

10-2 

8-6 

II-2 

19 
20 

•  • 

2-8 

4-0 
3-4 

5-5 
4'7 

7-3 
6-3 

9-5 

8-2 

10-4 

21 

3-o 

4-1 

5-4 

7-0 

9-0 

11-2 

22 

.. 

.. 

.. 

.. 

3-5 

4'7 

6-1 

7-8 

9-7 

23 

.. 

•  • 

3-i 

4-1 

5-4 

6-8 

8-5 

24 

•• 

2-7 

3-6 

4-7 

6-0 

7-5 

511 


TABLE  V. 


HEMLOCK   FLOOR   BEAMS   ONE    INCH   THICK,   FOR   FIRST-CLASS 

STORES. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams  Thicker  than   One  Inch,  see  Arts.   693  and  695. 


LENGTH 

DEPTH  OF  BEAM  (in  inches). 

BETWEEN 

BEARINGS 

(in  feet}. 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

8 

7-8 

9 

5-5 

7-8 

1O 

4-0 

5-7 

7-8 

11 

3-o 

4-3 

5-9 

12 

2-3 

3-3 

4-5 

6-0 

13 

1-8 

2-6 

3-6 

4-7 

6-2 

14 

•  • 

2-1 

2-8 

3-8 

4-9 

6-3 

15 

i-7 

2-3 

3-1 

4-0 

5-i 

6-4 

16 

1-9 

2-5 

3-3 

4-2 

5-2 

6-4 

17 

.. 

2-1 

2-8 

3-5 

4-4 

5-4 

6-5 

18 

•  • 

•  • 

1-8 

2-3 

2-9 

3  7 

4'5      5-5 

6-6 

19 

•• 

•• 

2-0 

2-5 

3-i 

3-8      4-7      5-6 

6-6 

20 

•• 

•• 

•• 

•• 

•• 

2-1 

2-7. 

3-3 

4-0      4-8 

5-7 

21 

1-9 

2-3 

2-8 

3'5 

4-1 

4*9 

22 

2-O 

2-5      3-o 

3'6     4-3 

0« 

2-2    1    2-6 

V2    !    V7 

44. 

I  •  O 

2  •  a 

2-8        V3 

OK 

1  .  o 

2-5        2-0    ' 

26 

2-2 

2-6 

512 


TABLE   VI. 


WHITE     PINE    FLOOR    BEAMS     ONE    INCH    THICK,    FOR    FIRST- 
CLASS    STORES. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams    Thicker  than   One  Inch,  see  Arts.   693  and  695. 


LENGTH 

DEPTH  OF  BEAM  (in  inches}. 

BETWEEN 

BEARINGS 

(in  feet}. 

8 

9 

1O 

11 

12 

13 

14 

15 

16 

17 

18 

8 

8-1 

9 

5-7 

8-1 

10 

4-1 

5-9 

11 

3'i 

4-4 

6-1 

12 

2-4 

3-4 

4-7 

6-2 

13 

1-9 

2-7 

3-7 

4-9 

6-4 

14 

2-2 

3-o 

3'9 

5'i 

6-5 

15 

•• 

i-7 

2-4 

3-2 

4-1 

5-3 

6-6 

16 

2-0 

2-6 

3-4 

4-3 

5-4 

6-7 

17 

2-2 

2-8 

3-6 

4'5 

5-6 

6-8 

18 

.. 

1-8 

2-4 

3-i 

3-8 

4-7 

5-7 

6-8 

•      < 

19 

2-0 

2-6 

3-2 

4-0 

4-8 

5-8 

6-9 

20 

•• 

•• 

•• 

•• 

2-2 

2-8 

3'4 

4-1 

5-o 

5-9 

21 

1-9 

2-4 

3-o 

3-6 

4-3 

5-1 

22 

2-  I 

2-6 

<2  -  I 

-i  .7 

4.  A 

23 

1-8 

2-2 

2-7 

3-3 

3-9 

*>4. 

2-O 

2-d 

2-Q 

3.  /i 

Q  K 

2-  I 

2-  ^ 

vo 

26 

1-9 

2-3 

2-7 

513 


TABLE    VII. 


SPRUCE    FLOOR    BEAMS    ONE    INCH    THICK,    FOR    FIRST-CLASS 

STORES. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams  Thicker  than  One  Inch,  see  Arts.  693  and  695. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet). 

DEPTH  OF  BEAM  (in  inches). 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

9 

6-9 

1O 

5-o 

7'i 

11 

3-8 

5-4 

7-3 

12 

2-9 

4-1 

5-7 

7-5 

13 

2-3 

3-2 

4-4 

5'9 

14 

1-8 

2-6 

3-6 

4'7 

6-2 

15 

•• 

2-  I 

2-9 

3-9 

5-0 

6-4 

16 

i-7 

2-4 

3-2 

4-1 

5-2 

6-5 

17 

2-0 

2-6 

3-4 

4-4 

5-5 

6-7 

18 

.. 

2-2 

2-9 

3-7 

4-6 

5-7 

6-9 

19 

.. 

.. 

1-9 

2-5 

3'i 

3-9 

4-8 

5-8 

20 

•• 



2-1 

2-7 

3-4 

4-1 

5-0 

6-0 

21 

1-8 

2-3 

2-9 

3-6 

4-3 

5-2 

6-2 

22 

.. 

.. 

.. 

2-0 

2-5 

3-1 

3-8 

4'5 

5'4 

23 

2-2        2-7 

3-3 

3-9 

4'7 

24 

I  -Q 

2-4 

2-9 

3-5 

4"  z 

•je 

2-  I 

2-6 

3'  * 

3-6 

26 

1 

1-9 

2-3 

2-7 

3-2, 

.  .     | 

TABLE   VIII. 


GEORGIA   PINE    FLOOR    BEAMS    ONE    INCH    THICK,    FOR    FIRST- 
CLASS    STORES. 

DISTANCE  FROM  CENTRES  (in  inches). 
For  Beams.  Thicker  than  One  Inch,  see  Arts,  693  and  695. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet}. 

DEPTH  OF  BEAM  (in  inches) 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

11 

6-3 

12 

4'9 

7-0 

13 

3-8 

5-5 

14 

3'i 

4.4 

6-0 

15 

2-5 

3-6 

4'9 

6-5 

16 

2-1 

2-9 

4-0 

5'4 

7-0 

It 

i-7 

2-4 

3-4 

4-5 

5-8 

18 

2-1 

2-8 

3-8 

4'9 

6-2 

19 

1-8 

2-4 

3'2 

4-2 

5'3 

6-6 

20 

•,• 

2-1 

2-7 

3'6 

4'5 

5*7 

21 

1-8 

2'4 

3'i 

3'9 

4'9 

6-0 

22 

2-1 

2-7 

3'4 

4-2 

5'2 

6.3 

23 

1-8 

2-3 

3-0 

3'7 

4-6 

5'5 

6-7 

24 

2-1 

2-6 

3'3 

4-0 

4.9 

5'9 

25 

•• 

1*8 

2-3 

2-9 

3-6 

4'3 

5-2 

6-2 

26 

•• 

•- 

•• 

-• 

•• 

2-1 

2-6 

3-2 

3-8 

4-6 

5'5 

515 


TABLE    IX. 


HEMLOCK  HEADERS  FOR  DWELLINGS,  OFFICE   BUILDINGS,  AND 
HALLS   OF   ASSEMBLY. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 
For  Tail  Beams  Longer  than   One  Foot,  see  Arts.  693  and  696a 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet). 

DEPTH  OF  HEADER  (in  inches). 

6 

7 

8 

9 

10 

11 

12 

13 

14 

5 

•33 

•  19 

•12 

.08 

6 

•58 

•33 

•21 

.14 

•10 

•07 

7 

•92 

•53 

•33 

•  22 

•  16 

•  ii 

•  09 

8 

1-37 

•79 

•50 

•33 

•23 

•17 

•13 

•  IO 

•08 

9 

i-95 

1-13 

•71 

.48 

•33 

•  24 

•  18 

•  14 

•  II 

10 

2-68 

i-55 

•  98 

•65 

•46 

•33 

•25 

•  19 

•15 

11 

^  .    '•'. 

2-06 

1-30 

•87 

•61 

•45 

•33 

•  26 

•20 

12 

2-68 

1-69 

1-13 

•79 

•  58 

•  43 

•33 

•  26 

13 

.. 

2-14 

i-44 

I-OI 

•73 

•55 

•43 

•33 

14 

•'   *  .  $ 

2.68 

1-79 

1-26 

•  92 

.69 

•53 

.42 

15 

... 

2-21 

i-55 

I-I3 

•85 

.65 

•5i 

16 

.  . 

.  . 

.  . 

2-68 

1-88 

i-37 

1-03 

•79 

•  62 

17 

..    . 

3.21 

2-26 

1-64 

1-24 

•95 

•75 

18 

. 

2-68 

i-95 

1-47 

I-I3 

-89 

19 

.. 

3-15 

2-80 

1-72 

i-33 

1-04 

2O 

•• 

3-67 

2-68 

2-OI 

1-55 

1-22 

516 


TABLE    X. 


WHITE    PINE    HEADERS    FOR    DWELLINGS,    OFFICE    BUILDINGS, 
AND   HALLS   OF   ASSEMBLY. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 
For   Tail  Beams  Longer  than  One  Foot,  see  Arts.  633  and  696. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet}. 

DEPTH  OF  HEADER  (in  inches). 

6 

7 

8 

9 

1O 

11 

12 

13 

14 

5 

•32 

.19 

•12 

.08 

6 

•56 

•32 

•20 

•  14 

•IO 

.07 

7 

.89 

•51 

•32 

•  22 

•15 

•  IT 

•  08 

8 

1.32 

•77 

•48 

•32 

•23 

•  16 

•12 

•IO 

•07 

9 

1-88 

1-09 

.69 

.46 

•32 

•24 

•18 

•  14 

•  ii 

1O 

2-59 

1.50 

•94 

•63 

•44 

•32 

.24 

•  19 

.16 

11 

1.99 

1-25 

-84 

•59 

•43 

•32 

•25 

•  20 

12 

2-59 

1-63 

I.Og 

•77 

.56 

.42 

•32 

•25 

13 

•  • 

.. 

2-07 

1-39 

•97 

•7i 

•53 

•41 

•32 

14 

.. 

2-59 

i-73 

1-22 

.89 

•  67 

•51 

•40 

15 

•• 

3-i8 

2-13 

1-50 

1-09 

•  82 

•63 

•50 

16 

.  . 

2-59 

1.82 

1-32 

•99 

•77 

.60 

17 

3-io 

2-18 

i-59 

1-19 

•92 

•72 

18 

2-59 

1-88 

1.42 

1-09 

•  86 

19 

3-04 

2-22 

1-67 

1-28 

I-OI 

20 

3-55 

2-59 

1.94 

1.50 

1.18 

517 


TABLE    XL 


SPRUCE    HEADERS    FOR    DWELLINGS,    OFFICE    BUILDINGS,   AND 
HALLS   OF  ASSEMBLY. 

THICKNESS  'OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 
For  Tail  Beams  Longer  than  One  Foot,  see  Arts.  693  and  696. 


LENGTH 

BETWEEN 

BEARINGS 

(in  feet). 

DEPTH  OF  HEADER  (in  inches). 

6 

7 

8 

9 

10 

11 

12 

13 

14 

5 

.27 

•15 

.10 

•06 

6 

.46 

•27 

•17 

•  ii 

.08 

7 

•73 

•42 

•  27 

.18 

•13 

.09 

8 

1.  10 

.63 

.40 

.27 

.19 

.14 

•  IO 

•08 

9 

1-56 

.90 

•57 

•33 

•27 

.19 

•15 

•  II 

•09 

1O 

2-14 

1.24 

.78 

•52 

•37 

•27 

•  20 

•15 

12 

11 

1-65 

1-04 

•7°   '     -49 

•36 

•27 

•  21 

•l6 

12 

2-14 

i-35 

.90 

.63 

.46 

•35 

•27 

•21 

13 

2-72 

1.72 

i-i5 

•  81 

•59 

•44 

•34 

•27 

14 

2-14 

i-43 

I-OI 

•73 

•55 

.42 

•33 

15 

•• 

•• 

2.63 

i-77 

1.24 

.90 

•  68 

•52 

.41 

16 

.  . 

3-20 

2-14 

1.50 

I-IO 

.82 

•  63 

•50 

17 

2-57 

i.  80 

1.32 

•99 

.76 

•  60 

18 

.. 

3-05 

2-14 

1-56 

r.zjf 

.90 

•71 

19 

2.52 

1.84 

1.38 

i.  06 

.84 

20 

2-94 

2.14 

1.61 

1-24 

•97 

518 


TABLE    XII. 


GEORGIA  PINE  HEADERS   FOR  DWELLINGS,  OFFICE   BUILDINGS, 
AND    HALLS   OF  ASSEMBLY. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 
For  Tail  Beams  Longer  than  One  Foot,  see  Arts.  693  and  696. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet). 

DEPTH  OF  HEADER  (in  inches). 

6 

7 

8 

9 

1O 

11 

12 

13 

14 

5 

•  16 

.09 

6 

•  27 

.16 

•  10 

•07 

7 

•44 

•25 

.16 

•  II 

.07 

8 

.65 

•38 

•  24 

•  16 

•  II 

•08 

9 

•93 

•54 

•34 

•23 

.16 

•12 

•09 

1O 

1-27 

•73 

.46 

•3i 

.22 

•  16 

•12 

-09 

11 

1-69 

.98 

•  62 

.41 

.29 

•  21 

•  16 

•  12 

•  10 

12 

2-  2O 

1-27 

.80 

•54 

•38 

•27 

•21 

•  16 

•T2 

13 

1.62 

!•  02 

.68 

.48 

•35 

•26 

.20 

•  16 

14 

2-02 

1-27 

•85 

.60 

•44 

•33 

•25 

•20 

15 

2-48 

1-56 

1-05 

•73 

•54 

•40 

•31 

•  24 

16 

.  . 

1.90 

1-27 

.89 

•  65 

•49 

•38 

•30 

17 

2-28 

1-52 

1.07 

.78 

•59 

•45 

•35 

18 

2-70 

i.  81 

1.27 

•93 

•70 

•54 

•42 

19 

2-13 

1.49 

1-09 

•  82 

•63 

.50 

2O 

2.48 

1.74 

1-27 

•95 

•73 

.58 

519 


TABLE    XIII. 


HEMLOCK   HEADERS   FOR   FIRST-CLASS   STORES. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 

For  Tail  Beams  Longer  than  One  foot,  see  Arts.  693  and  696. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet). 

DEPTH  OF  HEADER  (in  inches]. 

8 

9 

10 

11 

12        13 

14 

15 

16        17 

18 

5 

.28 

•19 

•13 

•  IO 

.07 

6 

.48 

•32 

•23 

•17 

.12 

.10 

•  07 

7 

•77 

•51 

.36 

•  26 

•2O 

.15 

•  12 

•10 

•  08 

8 

1.14 

•77 

•54 

•39 

.29 

•23 

.18 

.14 

•12 

.10 

•08 

9 

1.63 

1-09 

•77 

.56 

.42 

•32 

•25 

•  20 

•17 

.14 

•  II 

1O 

2-24 

1.50 

1.05 

•77 

•58 

•44 

•35 

.28 

( 
•23 

.19 

.16 

11 

2.98 

1.99 

1-40 

i  -02 

•77 

•59 

.46 

•37 

•30 

•25 

•  21 

12 

2-59 

1-82 

i-33 

LOO 

•77 

.60 

.48 

•39 

•32       -27 

13 

•« 

3-29 

2.31 

1-69 

1.27 

•97 

•77 

.61 

.50 

•4* 

•34 

14 

... 

,  ..    . 

2-89 

2-10 

1-58 

1-22 

.96 

.77  '     -62 

•5i 

•43 

15 

•• 

3-55 

2-59 

i-95 

1.50 

LI8 

•94       -77 

•63 

•53 

16 

.  . 

.  . 

.  . 

3-14 

2.36 

L82 

i-43 

i-i4       -93 

•77 

•64 

17 

3-77 

2.83 

2.1-8 

1-72 

i-37     1-12 

.92 

•77 

18 

•  • 

.. 

... 

3-36 

2-59 

2-04 

1-63     i-33 

1.09 

.91 

19 
20 

•• 

3-95 
4-61 

3-°5 

3-55 

2-39 
2-79 

1-92     1-56 
2-24     1.82 

1.28 
1-50 

1.07 
1-25 

520 


TABLE    XIV. 


WHITE   PINE    HEADERS   FOR   FIRST-CLASS   STORES. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 

For  Tail  Beams  Longer  than  One  Foot,  see  Arts.  693  and  696. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet}. 

DEPTH  OF  HEADER  (in  inches'). 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

5 

.27 

.18 

•13 

.09 

.07 

6 

•47 

•31 

.22 

.16 

.12 

.09 

•07 

7 

•74 

•50 

•35 

•25 

•19 

•15 

•12 

.09 

.07 

8 

i.ii 

•74 

.52       .38 

•  28 

.22 

•17 

•  14 

.11 

•09 

.08 

9 

i-57 

1-05 

-74 

•54 

.41 

•31 

•25 

.20 

.16 

•13 

•  ii 

10 

2.16 

1-46 

1.02 

•74 

.56 

•43 

•34 

.27 

•  22 

.18 

•15 

11 

2-87 

i-93 

1-35 

•99 

•74 

•57 

•45 

.36 

.29 

•  2*4 

•20 

12 

2.50 

1.76 

1.28 

.96 

•74 

.58 

•47 

-38 

•31 

•  26 

13 

•  • 

3-i8 

2-23 

1.63 

1.22 

•94 

•74 

•59 

.48 

•40 

•33 

14 

2.79 

2-03 

i-53 

1.18 

.92 

•74 

.60 

•  50 

.41 

15 

3-43 

2.50 

1.88 

i-45 

1-14 

.91 

.  -74 

.61 

•5i 

16 

.  . 

3-03 

2-28 

1.76 

1-38 

i-ii 

•90 

•74 

•  62 

17 

3-64 

2-73 

2.  II 

i-66 

i-33 

I-  08 

•  89 

•74 

18 

•  • 

4-32 

3-25 

2-50 

i-97 

1-57 

1-28 

1-05 

•  88 

19 

3-82 

2-94 

2.31 

1-85 

1.50 

1.24 

1-03 

20 

•- 

•• 

4-45 

3-43 

2-70 

2-16 

1.76 

1-45 

1.  21 

1 

521 


TABLE    XV. 


SPRUCE   HEADERS   FOR   FIRST-CLASS   STORES. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 

for  Tail  Beams  Longer  than  One  Foot,  see  Arts.  693  and  696. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet). 

DEPTH  OF  HEADER  (in  inches). 

8 

0 

10 

11 

13 

13 

14 

15 

16 

17 

18 

5 

.22 

•15 

•IO 

.08 

6 

•39 

.26 

•  18 

•13 

•10 

.08 

7 

•61 

.41 

•29 

•21 

.16 

.12 

•IO 

.08 

8 

•92 

.61 

•43 

•31 

.24 

.18 

•  14 

•  II 

•  09 

•  08 

9 

1-30 

.87 

.61 

•45 

•34- 

•26 

•  20 

.16 

•13 

•  II 

•  09 

10 

1.79 

1-20 

.84 

.61 

.46 

•35. 

.28 

•  22 

.18 

•15 

•  12 

11 

2-38 

I-  60 

I-I2 

•  82 

•61 

•47 

•37 

•30 

•24 

•  2O 

•17 

19 

3-09 

2-07 

i-45 

1-06 

•  80 

•  61 

.48 

•39 

•31 

.26 

.22 

13 

2.63 

1-85 

i-35 

1.  01 

.78 

.61 

•49 

.40 

•33 

•  27 

14 

•  • 

3-29 

2.31 

i-68 

1-26 

•97 

•77 

•  61 

•50 

.41 

•34 

15 

•• 

2-84 

2.07 

1.56 

1-20 

•94 

•75 

•  61 

•5i 

•42 

16 

3-45 

2-51 

1.89 

i-45 

1-14 

•92 

•74 

•  61 

•5i 

17 

3-02 

2-27 

1.74 

i-37 

I-IO 

.89 

•74 

•61 

18 

.. 

.. 

.. 

3-58 

2-69 

2-07 

1-63 

1.30 

I-  06 

•87       -73 

10 

•  • 

•• 

•  • 

4-21 

3-i6 

2.44 

1-92 

i-53 

1-25     1.03 

•  86 

20 

•5.60 

2-84 

2-23 

i-79 

i-45 

1-20 

I-  00 

522 


TABLE    XVI. 


GEORGIA   PINE   HEADERS  FOR  FIRST-CLASS  STORES. 

THICKNESS  OF  HEADER  (in  inches)  FOR  TAIL  BEAMS  ONE  FOOT  LONG. 
For  Tail  Beams  Longer  than  One  Foot,  see  Arts.  693  and  696. 


LENGTH 

BETWEEN 

BEARINGS 
(in  feet}. 

DEPTH  OF  HEADER  (in  inches). 

• 

9 

10 

n 

13 

13 

14 

15 

16 

17 

18 

5 

.13       -eg 

.06 

6 

•23  :    -15 

•  II 

.08 

7 

•36  !    -24 

.46 

.12 

•09 

.07 

8 

•54 

•36 

•25 

•I9 

•  14 

•  II 

.08 

9 

•77  !     -52 

•36 

•  26 

.20 

•15 

.12 

.IO 

•  08 

10 

i.  06 

•71 

•50 

•36 

•27 

•21 

•17 

•13 

•  II 

•  09 

11 

1-41 

•95 

.67 

.48 

•36 

.28 

•  22 

.18 

.T4. 

•  12 

•10 

12 

1.83 

1-23 

•  86 

•63 

•47 

•36 

•29 

•23 

.I9 

•15 

•13 

13 

2-33 

1-56 

I-IO 

.80 

•60 

.46 

.36 

.29 

•24 

.19 

.16 

14: 

2-91 

i-95 

1-37 

I-OO 

•75 

•58 

•45 

.36 

•30 

.24 

•  20 

15 

2-40 

1.69 

1-23 

.92 

•71 

•56 

•45 

•36 

•30 

•25 

16 

.  . 

2.91 

2.05 

1-49 

I-  12 

•  86 

•68 

•54 

•44 

•36 

•30 

17 

3-49 

2-45 

1.79 

i-34 

1-03 

•  81 

.65 

•53 

•44 

•36 

18 

.. 

2-91 

2-12 

i-59 

1-23 

•97 

•77 

•63 

•52 

•43 

19 

.. 

3-43 

2-50 

1.88 

1.44 

1.14 

.91 

•74 

•61 

•5i 

30 

•• 

2-gi 

2-19 

1-69 

1-33 

i.  06 

.86 

•7i 

•59 

1 

523 


TABLE    XVII. 


ELEMENTS   OF   ROLLED-IRON   BEAMS. 
See  Art.   697. 


NAME. 

i 

Q 
1! 

** 

WEIGHT  PER 
YARD. 

*  = 

BREADTH. 

AVERAGE 
THICKNESS 
OF  FLANGE. 

i 

THICKNESS 
OF  WEB. 

6, 

dt 

X 

~oT 

ii  i 
s| 

Pittsburgh  . 

3 

21 

2.316 

•359 

.191 

2.125 

2.281 

3.109 

Pittsburgh  . 

3 

27 

2.516 

-359 

•391 

2.  125 

2.281 

3-559 

Phoenix  .  .  . 

4 

18 

2. 

.278 

.2 

T.8 

3.444  ;     4.539 

Trenton  .  . 

4 

18 

2. 

.29 

.187 

I.8I3 

3.42     i     4.623 

Pottsville.  . 

4 

18 

2.125 

.271 

.187 

1.938 

3.458   ;     4.655 

Paterson  .  . 

4 

18 

2.25 

.281 

.156 

2.094 

3.438        4-909 

Pittsburgh  . 

4 

24 

2.481 

.328 

.231 

2.25 

3.344          6.221 

Pittsburgh  . 

4 

30 

2-631 

.328 

.381 

2.25 

3.344  i     7-021 

Pottsville.  . 

4 

30 

2.25 

•5 

•25 

2. 

3- 

7-5 

Buffalo  

4 

30 

2,75 

•4 

•25 

2-5 

3-2 

7.840 

Paterson  .  . 

4 

30 

2-75 

•4 

•  25 

2-5 

3-2 

7.840 

Phoenix  .  .  . 

4 

30 

2-75 

•4 

•25 

2-5 

3  •  2 

7.840 

Trenton.  .  . 

4 

30 

2-75 

-25 

2-5 

3-2 

7.840 

Paterson  .  . 

4 

37 

3-               -456 

.312 

2.688 

3.088 

9.404 

Trenton.  .  . 

4 

37 

3- 

•456 

.312 

2.688 

3.o8S 

9.404 

Buffalo  

5 

3^ 

2-75 

•35 

•25 

2-5 

4-3 

12.082 

Paterson  .  . 

5 

30 

2-75 

-35 

•25 

2-5 

4-3 

12.082 

Phoenix  .  .  . 

5 

30 

2-75 

•35 

-25 

2-5 

4-3 

12.082 

Trenton..  . 

5 

30 

2-75 

•35 

2-5 

4-3 

12.082 

Pottsville.  . 

5 

30 

3.062 

•3ii 

•25 

2.812 

4-378 

12.232 

Pittsburgh. 

5 

30 

2.725 

•375 

•  225 

25 

4-25 

12.393 

Pittsburgh  . 

5 

39 

2.905 

•375 

•4^5 

2-5 

4-25 

14.268 

Phoenix  .  .  . 

5 

36 

3- 

•389 

-3 

2.7 

4.222 

I4-3I7 

Paterson  .  . 

5 

40 

3- 

.438 

•333 

2.  667 

4.125 

15-650 

Trenton..  . 

5 

40 

3- 

•454 

.312 

2.688 

4.092 

15-902 

Pottsville.  . 

5 

40 

3-125 

•434 

.312 

2.813 

4.132 

16.015 

Phoenix  .  .  . 

6 

40 

2  75 

•5 

•25 

2-5 

5- 

23-458 

Buffalo  

6 

40 

3- 

•454 

•25 

2.75        5.091 

23-761 

Paterson  .  . 

6 

40 

3- 

•454 

-25 

2-75          5-°9i  '   23.761 

Trenton.  .  . 

6 

40 

3- 

•454 

•25 

2-75          5-°9*      23.761 

524 


TABLE    XVII.— (Continued^ 


ELEMENTS   OF   ROLLED-IRON   BEAMS. 
See  Art.  697. 


i 

s  . 

E 

§11 

t/2 

«  g 

X 

HI 

NAME. 

S 

1  * 

11  § 

•^    rj 

W  ^  rv' 

w  fa 

bt 

d, 

1 

s! 

w 

PQ 

<^o 

H  ° 

Pottsville.. 

6 

40 

3-375 

•4 

•25 

3.125 

5-2 

24-T33 

Pittsburgh. 

6 

4°i 

3-237 

•437 

•237 

3- 

5-125 

24.613 

Pittsburgh. 

6 

54" 

3.462 

•437 

.462 

3- 

5-125 

28.663 

Buffalo  

6 

50 

3-25 

•532 

312 

2.938 

4-935 

29.074 

Pottsville.  . 

6 

50 

3-437 

.500 

.312 

3.125 

5.080 

29.314 

Phoenix  .  . 

6 

50 

3-5 

.492 

•31 

3-19 

5.016 

29-451 

Paterson  .  . 

6 

SO' 

3-5 

•5 

•3 

3-2 

5- 

29.667 

Trenton.  .  . 

6 

50 

3-5 

•  5 

•3 

3-2 

5- 

29.667 

Phoenix  .  .  . 

7 

55 

3-5 

•484 

•35 

3.15 

6.032 

42.430 

Pottsville.  . 

7 

55 

-510 

.312 

3-25 

5.980 

43-897 

Trenton  .  .  . 

7 

55 

3-75 

•493 

•  3 

3-45 

6.014 

44.652 

Pittsburgh. 

7 

54 

3.604 

•  562 

.229 

3-375 

5-875 

45-983 

Buffalo  

.      7 

60 

3-5 

•54 

•375 

3-125 

5-92 

46.012 

Paterson  .  . 

7 

60 

3-5 

•54 

•375 

3-125 

5.92 

46.012 

Phoenix  .  .  . 

7 

69 

3-687 

.476 

•56 

3.127 

6.048 

47-739 

Pottsville.. 

7 

65 

3.625 

•596 

•375 

3-25 

5-8oS 

50.553 

Paterson  .  . 

6 

90 

5- 

.667 

•  5 

4-5 

4-667 

51.881 

Trenton.  .  . 

6 

90 

5- 

.667 

4-5 

4.667 

51.881 

Pittsburgh. 

7 

75 

3-9°4 

.562 

.529 

3-375 

5-875 

54.558 

Buffalo  

8 

65 

3-5 

•56 

•375 

3-125 

6.880 

64.526 

Paterson  .  . 

6 

1  20 

5-5 

•789 

•75 

4-75 

4.422 

64.773 

Phoenix  .  .  . 

8 

65 

4- 

.478 

•38 

3-62 

7.044 

65.232 

Trenton.  .  . 

6 

120 

5-25 

.892 

.625 

4-625 

4.216 

65.618 

Pottsville. 

8 

65 

4- 

•543 

.312 

3.688 

6.914 

69  .  089 

Paterson  .  . 

8 

65 

4- 

-554 

•3 

3-7 

6.892 

69.729 

Trenton.  .  . 

8 

65 

4- 

=  554 

•3 

3-7 

6.892 

69.729 

Pittsburgh. 

8 

66 

3.806 

•593 

•  306 

3-5 

6.813 

70.153 

Pottsville.. 

8 

80 

4.187 

•542 

•  5 

3-687 

6.916 

77.007 

Phoenix  .  .  . 

8 

81 

4.125 

•556 

3-6i5 

6.888 

77-552 

Buffalo  

9 

70 

3-5 

.500 

•437 

3-063 

8.000 

81-937 

525 


TABLE     XVII.— (Continued.) 

ELEMENTS   OF   ROLLED-IRON   BEAMS. 
See  Art,  697. 


NAME. 

d  =  DEPTH. 

g 

el 

s> 

M 

*s= 
BREADTH. 

AVERAGE 
THICKNESS 
OF  FLANGE. 

THICKNESS 
OF  WEB. 

*, 

4 

V 
•<f 

ii  i 
sl 

Trenton  .  .  . 

8 

80 

4-5 

.606 

-375 

4.125 

6.788 

84.485 

Paterson  .  . 

8 

80 

4-5 

.610 

-37 

4-13 

6.780 

84-735 

Pottsville.. 

o. 

70 

4.125 

•483 

•375 

3-75 

8.034 

88-545 

Pittsburgh. 

8 

105 

4-293 

-593 

-793 

3-5 

6.813 

90.932 

Phoenix  .  .  . 

9 

70 

3-5 

.660 

•3i 

3-19 

7.680 

92.207 

Paterson  .  . 

9 

70 

3-5 

.672 

•3 

3-2 

7.656 

92.958 

Trenton.  .  . 

9 

70 

3-5 

.672 

•  3 

3-2 

7-656 

92.958 

Pittsburgh. 

9 

70-V 

4.012 

-625 

.262 

3-75 

7-75 

98.265 

Pottsville.  . 

9 

90 

4-5 

.501 

•  562 

3-938 

7.998 

105.480 

Phoenix  .  .  . 

9 

84 

4- 

.667 

•4 

3-6 

7.667 

107.793 

Buffalo  

9 

90 

4- 

.643 

•5 

3-5 

7-714 

109.117 

Paterson  .  . 

9 

85 

4- 

.697 

•384 

3.616 

7.605 

110.461 

Trenton.  .  . 

9 

85 

4- 

.707 

•375 

3-625 

7-586 

in  .  124 

Pittsburgh. 

9 

99 

4-329 

.625 

•579 

3-75 

7-75 

117-523 

Pottsville.. 

ioi 

90 

4-25 

.578 

•437 

3-8i3 

9-344 

150.763 

Pittsburgh. 

10 

90 

4-325 

.718 

•325 

4- 

8-563 

151-123 

Buffalo  

ioj 

90 

4-437 

•55i 

•437 

9-397 

151-436 

Trenton..  . 

9 

125 

4-5 

•937 

-57 

3-93 

7-125 

i54-9T7 

Pittsburgh. 

9 

135 

4-931 

.812 

•744 

4.187 

7-375 

159-597 

Pittsburgh. 

10$ 

94* 

4-534 

.625 

.409 

4.125 

9-25 

165-327 

Pottsville.. 

10} 

105 

4-562 

•575 

.562 

4- 

9-350 

167.624 

Trenton..  . 

10} 

90 

4-5 

.683 

.312 

4.188 

9.434 

168.154 

Pittsburgh  . 

9 

150 

5-098 

.812 

.911 

4.187 

7-375 

169.742 

Buffalo.  .  .  . 

10* 

105 

4-5 

•  656 

-5 

4- 

9.187 

175-645 

Phoenix  .  .  . 

io| 

105 

4-5 

.724 

•44 

4.06 

9.052 

183.164 

Pittsburgh. 

10 

135 

4-775 

.718 

•775 

4- 

8-563 

188.623 

Phoenix  .  .  . 

9 

150 

5-375 

1.005 

.6 

4-775 

6.99      190.630 

Paterson  .  . 

io.V 

105 

4-5 

•795 

•375 

4.125 

8.909    191.040 

Trenton.  .  . 

roi 

105 

4-5 

•795 

•375 

4-125 

8.909    191.040 

Phoenix  .  .  . 

H>* 

135 

4-875 

.614 

.81 

4-065 

9.272 

200.263 

525  a 


TABLE    1KM\\.— (Continued.) 

ELEMENTS      OF      ROLLED-IRON      BEAMS. 
See  Art.  697. 


NAME. 

| 

W 

Q 
1! 
> 

WEIGHT  PER 
YARD. 

b  = 

.  BREADTH. 

AVERAGE 
THICKNESS 
OF  FLANGE. 

i 
THICKNESS 
OF  WEB. 

ft 

< 

N 

V 

nf 
sl 

<w 

Pittsburgh. 

10  V 

i35 

4.920 

.625 

•795 

4.125 

9  25 

202.564 

Paterson  .  . 

10:} 

135 

5- 

•945 

•47 

4-53 

8.609 

241.478 

Trenton.  .  . 

io| 

135 

5- 

•945 

•47 

4-53 

8.609 

241.478 

Pottsville.  . 

12 

-25 

4.562 

.800 

•5 

4.062 

10.400 

276.162 

Pittsburgh. 

12 

126 

4-638 

.781 

•5i3 

4.125 

10.438 

276.946 

Phoenix  .  .  . 

12 

125 

4-75 

•777 

•49 

4.26 

10.446 

279.351 

Buffalo.  .  .  . 

I24L 

125 

4-5 

•797 

-5 

4- 

10.656 

286.019 

Paterson  .  . 

12\ 

125 

4-79 

.768 

.48 

4-3i 

10.714 

292.050 

Trenton.  .  . 

TO1 

I24- 

125 

4-79 

.780 

•47 

4-32 

10.690 

293.994 

Pittsburgh. 

12 

1  80 

5.088 

.781 

•963 

4  125 

10.438 

341.746 

Pottsville.. 

12 

170 

5- 

i.oS6 

.625 

4-375 

9.828 

373-907 

Phoenix    .  . 

12 

170 

5-5 

I.OTO 

•59 

4.9: 

9.980 

385.284 

Paterson  .  . 

I2i 

170 

5-5 

.985 

.6 

49 

10.28 

398.936 

Trenton.  .  . 

12  fs 

170 

5  5 

.981 

.6 

49 

10.351 

402.538 

Buffalo  

I2f 

1  80 

5-375 

1.089 

.625 

4-75 

10.072 

418.945 

Buffalo  

15 

150 

4-875 

.761 

•  562 

4-3*3 

13-477 

491.307 

Paterson  .  . 

i5-,36- 

150 

5- 

•731 

•56 

4.440 

13-725 

502.883 

Phoenix  .  .  . 

15   ' 

150 

4-75 

.882 

-5 

4-25 

13-235 

514.870 

Pottsville.  . 

15 

150 

5.062 

.822 

•5 

4.562 

13-356 

517.948 

Trenton..  . 

ISA 

150 

5- 

.822 

-5 

4-5 

I3-542 

528.223 

Pittsburgh  . 

15 

150 

5  030 

.875 

.468 

4.562 

13-25 

530.343 

Pittsburgh. 

15 

195 

5-330 

.875 

.768 

4.562 

13-25 

614.718 

Pittsburgh. 

15 

2OI 

5-545 

.031 

.670 

4-875 

12    938 

679.710 

Phoenix  .  .  . 

15 

2OO 

5-375 

.085 

-65 

4-725 

12.830 

680.  146 

Buffalo.... 

15 

20O 

5-375 

.118 

.625 

4-75 

12.763 

688.775    ! 

Paterson  .  . 

15* 

200 

5-5 

.048 

-65 

4.85 

13.028 

692.166 

Pottsville.  . 

15 

2OO 

5.687 

.04x5 

.625 

5.062 

1  2  .  902 

693.503 

Trenton.  .  . 

i5i 

2OO 

5-75 

.060 

.6 

5-15 

I3-004 

714.205 

Pittsburgh. 

15 

240 

5-805 

.031 

930 

4-875 

12.938 

752.835 

! 

525  b 


TABLE   XVIII. 


ROLLED-IRON   BEAMS   IN   DWELLINGS,    OFFICE   BUILDINGS,  AND   HALLS 

OF    ASSEMBLY. 

DISTANCES  FROM  CENTRES  (in  feet}. 
See  Arts.   694,  698  and  7OO. 


NAME. 

DEPTH. 

WEIGHT  PER 
YARD. 

LENGTH  (in  feet)  BETWEEN  BEARINGS. 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

22 

23 

Pittsburgh.. 

3 

21 

3.62 

2.26 

L50 

Pittsburgh.. 

3 

27 

-.14 

2.58 

1.71 

.18 

Phoenix  

4 

18 

5.32 

3.33 

2.22 

.54 

.11 

Trenton..  .  . 

4 

18 

i-42 

1.39 

2.26 

•  57 

•  T4 

Pottsville... 

4 

18 

5.45 

2.28 

.59 

.14 

Paterson  .  .  . 

4 

18 

3.61 

2.4O 

.67 

.21 

Pittsburgh.. 

4 

24 

* 

[•57 

3.°4 

.12 

.53 

1.13 

Pittsburgh.. 
Pottsville... 

4 
4 

30 
3° 

'• 

•  39 

1.27 

Buffalo  

4 

30 

5^6 

3.83 

.67 

.93 

L43 

Paterson  .  .  . 

4 

3° 

5.76 

3.83 

.67 

•93 

!-43 

Phoenix.... 

4 

3° 

5.76 

383 

.67 

•  93 

L43 

Trenton...  . 

4 

3° 

5.76 

3.83 

.67 

•93 

Paterson  .  .  . 

4 

37 

4.60 

3.20 

•  31 

1.71 

.30 

Trenton.... 

4 

37 

4.60 

3.20 

•  3° 

Buffalo  

5 

3° 

5.95 

4.16 

3.01 

2.24 

•  71 

•  33 

Paterson  .  .  . 

5 

3° 

5-95 

4.16 

3.01 

2.24 

•  71 

•  33 

Phoenix.... 

5 

3° 

5-95 

4.16 

3.01 

2.24 

«7l 

.33 

Trenton  

5 

3° 

5-95 

4.16 

3.01 

2.2^ 

•  71 

•  33 

Pottsville..  . 

5 

3° 

.. 

.. 

6.  02 

4.21 

3.05 

2.27 

•  73 

•  35 

Pittsburgh. 
Pittsburgh. 

5 
5 

30 
39 

6.10 
7.02 

4.26 

4.90 

3.°9 
3.55 

i-.i: 

.76 
.01 

•  37 
•  56 

1.23 

Phoenix  .  . 

5 

36 

7-°5 

4.92 

3.57 

2.66 

.03 

.58  1.24 

Paterson  .  . 

5 

40 

\\ 

5.38 

3.9° 

2.90 

.21 

.721.36 

Trenton  .  .  . 

5 

40 

•• 

•• 

.• 

5.47 

3.96 

2.95 

.25 

•75  1.38 

Pottsville.  . 

5 

40 

5.51 

3-99 

2.97 

2.27 

.76  L39 

Phoenix... 
Buffalo  .  .  . 

6 
6 

40 
40 

'* 

•• 
'* 

•• 

8.  ii 

8.22 

5.89 
5-97 

4.40 
4.46" 

3-37 

.63  2.09 
.662.11 

1.68 
1.70 

;-j 

Paterson  .  . 

6 

40 

8.22 

5-97 

4.46 

3.4' 

.662.11 

i.7c 

r^s 

Trenton  .  .  . 

6 

40 

8.22 

5-97 

4.46 

3.41 

.662.11 

1.70 

!.•}£ 

1 

526 


TABLE    XVIIL— (Continued.} 


ROLLED-IRON   BEAMS   IN   DWELLINGS,    OFFICE   BUILDINGS,  AND    HALLS 

OF  ASSEMBLY. 

DISTANCES  FROM  CENTRES  (in  feet). 
See  Arts.   694,  698  and  7OO. 


NAME. 

DEPTH. 

WEIGHT  PER  | 
YARD. 

LENGTH  (in  feet)  BETWEEN  BEARINGS. 

6 

7 

8 

9 

1O 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

21 

22 

23 

Pottsville... 
Pittsburgh.. 

6 
6 

40 

•  • 

5.ob 
6.18 

4.53 
-1.62 

3.47 
3.54 

2.71 

2.76 

.15 

.IQ 

•  73 
.76 

.44 

i  Pittsburgh.. 

6 

S4 

7  18 

4.10 

3.20 

•  54 

.04 

66 

1  .36 

Buffalo  .... 

6 

50 

.. 

7.30 

5.45 

4.17 

3.26 

•  S8 

.08 

.69 

1.39 

Pottsville... 

6 

5° 

i   -6 

5  5° 

4.21 

7     •>« 

.61 

.10 

.71 

1  .40 

Phoenix..  .  . 

6 

5° 

o  (r, 

Paterson  .  .  . 

6 

so 

7-45   S.S7 

4  26 

3.33 

2.64 

2.12 

•  73 

1.42 

Trenton  

6 

50 

7.45 

"•57 

4.26 

2  .64  2.12 

.73 

1.42 

Phoenix  

7 

55 

8.00 

6.n 

t-79 

3.8i!3.c8 

•  Si 

2.07 

.72 

L45 

Pottsville  .. 

7 

55 

•  • 

8.28 

6.35 

4-97 

3-95 

3-19 

.60 

2.15 

•79 

i-5C 

Trenton  
Pittsburgh.. 
Buffalo  

7 
7 
7 

55 
54 
60 

.. 

.. 

8.43 

8.68 

6.46 
6.6f 
6.6=; 

5.C-5 

5.21 

3.20 

4.0213.24 

4-I53.35 
4.1313.33 

.65 
•  73 
.72 

2.19 

2.  2O 
2.25 

.82 
.88 

87 

1.53 
1.58 

.29 

.34 
.32 

Paterson.  .  . 

7 

60 

6.6= 

5.20 

4.!3  3-33 

.72 

2.25 

.87 

1  .57 

.32 

Phoenix.... 

7 

69 

8.98 

6.88 

5.38 

4.27 

3-44 

.81 

2.31 

•  92 

i  .61 

•  36 

Pottsville... 

7 

65 

7-31 

5.7* 

4.54 

3.67 

2.09 

2.47 

.06 

1.72 

>4f 

i  Paterson  .  .  . 
Trenton  
Pittsburgh.. 
Buffalo  

6 
6 

7 
8 

90 
90 

65 

'• 

7-44 
7.87 
9-37 

-.8l 

5.8i 
5.i6 

7-34 

4'.6\ 

4.89 
5.84 

3.71 

3-71 
3-94 
4.72 

3-02 
3.02 
3.22 

3.86 

2.48 
2.48 

1% 

.05 

.05 

.21 

.67 

1.71 
L71 
1.85 
2.24 

•  44 
•  44 
.56 
.90 

.32 
.62 

1.39 

Paterson  .  .  . 
Phoenix  
Trenton  
Pottsville... 
Paterson.  .  . 

6 
8 
6 

8 
8 

120 

65 
120 

•• 

;; 

••- 

9.28 
9-47 
9-4C 
10.04 
10.14 

7.23 
7.42 
7-33 
7.87 
7-94 

5-73 
5.91 
S.oi 

6.27 
6.33 

4.6i 

4.77 
4.67 
5.07 

3.75 

3^0 
4.15 
4.19 

3.08 
3.23 
3.12 

3-43 
3.46 

.55 
.'87 

2.12 
2.27 
2.15 
2.41 

2.44 

.7* 

.92 

.05 
.07 

•  501.27 
.64    .41 
.52    .29 
.75,    -So 
•77[    .5* 

1.31 

Trenton  
Pittsburgh.. 
Pottsville... 
Phoenix  
Buffalo  

8 
8 
8 
8 
9 

65 

66 
80 
81 
70 

.. 

10.14 

IO.2O 

7-94 
7-99 

3  75 

6.33 
6.36 
6.97 
7.02 
7.45 

5." 
5-'4 
5.63 
5.67 
6.03 

4.19 

4-21 

4.61 

4.64 

4.94 

3.46 
3.48 
3.8i 
3.83 
4-09 

.89 
•  91 
3-i8 

3.20 

3.42 

2.44 

2:67 
2.69 

.07 
.08 
.26 
.28 
•  45 

•  77 
.77 

.09 

'S 

!so 

1.31 

1.42 
1.43 
i.55 

•• 

•••. 

8.  8  1 

9-35 

527 


TABLE    XVIII.— (Continued^ 


ROLLED-IRON   BEAMS     IN     DWELLINGS,     OFFICE     BUILDINGS, 
AND    HALLS   OF   ASSEMBLY. 

DISTANCES  FROM  CENTRES  (in  feet). 
See  Arts.  694,  698  and  7OO. 


0 

9* 

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527  a 


TABLE    l£Vl\\.— (Continued.) 


ROLLED-IRON   BEAMS    IN     DWELLINGS,     OFFICE    BUILDINGS, 
AND    HALLS   OF   ASSEMBLY. 

DISTANCES  FROM  CENTRES  (in  feet). 
See  Arts.  694,  698  and  7OO. 


LENGTH  (infect)  BETWEEN  BEARINGS. 

O 
M 

§ 

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30 

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3 

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FH 

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M 

FH 

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co  co  co  o?lN        c?  c?  c?  -c?co        S.£.R.S.oo         toS^OLOLO        LoSoSS        8885- 

fff  .  .    B¥S¥g    ss?fS   -j-fsp-f  jo.,.,.,.,,  S-jf., 

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527  b 


TABLE    XIX. 

ROLLED-IRON   BEAMS   IN   FIRST-CLASS   STORES. 

DISTANCES  FROM  CENTRES  (in  feet). 
See  Arts.  694,  697,  699  and  7OO. 


NAME. 

X 

K 

n 

WEIGHT  PER 
YARD. 

LENGTH  (infect)  BETWEEN  BEARINGS. 

5 

6 

7 

8 

9 

10 

11 

1«|  13    14 

15 

16   17 

18    19   2O 

Pittsburgh  

3 

21 

3.68 

2.12 

1.33 

Phoenix  

SB 

e    -38 

Trenton  
Pottsville  

4 

18 
18 

0 

5.48 

3.17 

[.99 

1.32 

18 

e    82 

Pittsburgh  

4 

24 

4,26 

2.67 

1.78 

1.24 

Pittsburgh  
Pottsville  

4 

4 

3° 
3° 

4.81 

3.01 
3.22 

2.OI 
2.15 

1.40 
1.50 

Buffalo  

4 

30 

5-37 

3-37 

2.25 

1-57 

1.14 

Paterson  

4 

30 

5.37 

3.37 

2.2S 

1-57 

.14 

Phoenix  

4 

30 

5.37 

3.37 

2.2.S 

1-57 

.14 

4 
4 

30 
37 

" 

5-37 

3-37 
4.04 

2.25 

2.6q 

i.57 
1.88 

.14 

.36 

Paterson  

Trenton  

"  fin 

i  88 

36 

Buffalo  

5 

S 

30 
30 

5.21 
5.21 

3.48 
3.48 

2-43 
2.43 

.77 
•  77 

1.32 
1.32 

Paterson  

Phoenix  

T      4.8 

Trenton  
Pottsville  

5 
5 

30 
30 

;; 

5.21 

5.27 

3.48 

3.52 

2.43 
2-47 

•  77 
•79 

1.32 
1-34 

Pittsburgh  
Pittsburgh  

5 

S 

30 

39 

•• 

5-34 

3.57 
4.11 

2.50 

.81 
.08 

!-35 

.IQ 

Phoenix 

nfi 

OQ 

rv\ 

z: 

Paterson  

6*17 

* 

6*86 

A      Cfi 

Pottsville  

40 

6.QI 

.61 

3.23 

2.34 

1.75 

Phoenix  

£ 

o    58 

Q8 

1.55 

i  23 

Buffalo  

6 

86 

4  81 

2    6l 

Paterson  
Trenton 

6 
6 

4° 
40 

.86 
.86 

4.81 
4.8i 

3-49 
3-49 

2.61 

2.6l 

.OO 
.OO 

i.57 
1-57 

1.22 
1.22 

528 


TABLE  XIX.— (Continued^ 


ROLLED-IRON   BEAMS   IN   FIRST-CLASS   STORES. 

DISTANCES  FROM  CENTRES  (in  feet}. 
See  Arts.  694,  697,  699  and  7OO. 


NAME. 

DEPTH. 

WEIGHT  PER  1 
YARD. 

LENGTH  (in  feet}  BETWEEN  BEARINGS. 

5 

6 

7 

8 

9 

10 

11 

12 

13 

11 

15 

16   17 

18 

19 

*o 

Pottsville  
Pittsburgh  

6 
6 
6 
6 

40 

54 

!   8R 

3  55 

->  66 

.27 

.23 
•24 

.25 
.26 
.26 
.81 
.88 

.91 
•97 
•97 
•97 
•  °3 

.16 
.19 
.19 

[78 

.73 
.81 
.77 
.98 
3.01 

3.01 

3.02 

3.31 
3-33 
3-54 

1.48 

1.54 

i!6i 

i!6i 
1.66 

1-77 
i.79 
1.79 

2.23 
2.30 
.26 
.44 
.46 

.46 
.48 
•  71 
.73 
.9° 

1.27 

•29 
•  34 
.33 
.33 
.37 

.46 
.48 
.48 
•  57 
•89 

.84 

£ 

.02 
.04 

.04 
.06 

.25 
.26 

•  41 

:£ 

•  71 

•  71 
•  72 
.88 
•  89 

.02 

1.33 

i.28 
•  35 
•30 
•  43 
.44 

•  44 
•  45 
•59 

.00 

.70 

:1 

•• 

•• 

7-11 
8.27 
R    10 

4-98 

3.62 
4.21 

4  27 

2.71 
3.20 

2.41 
~  .45 

'.88 
.92 
•93 

•94 

!?6 
2.82 
2.92 

2.97 
3.06 

3-°5 
3.05 
3.16 

3.36 
3.42 
3-42 
3.62 
4-30 

4.26 
4-35 
4.32 
4.61 
4.65 

4.65 
4.68 
5.13 
5.17 
5.48 

.29 

.50 
.52 
•  54 

!s6 
•  24 
•  32 

•  36 
•  44 
•  43 
•  43 
•  52 

.67 
.72 
.72 
2.88 
3-43 

3.39 
3.47 
3.43 
3.68 

3^73 
4.09 
4.12 

4-37 

Buffalo  

Pottsville  

6 

6 
6 
6 
7 
7 

7 
7 
7 
7 
7 

7 
6 
6 

6 
8 
6 
8 
8 

8 
8 
8 
8 
9 

50 

50 
50 
So 
55 
55 

55 

& 

60 
69 

65 
9° 
90 

75 
65 

120 
65 
120 

65 
65 

65 

66 
80 
81 
7° 

•• 

•• 

•• 

8.47 

1:11 

8.57 

5-93 

5.96 
6.00 
6.00 
8.60 
8.90 

9.06 
9-33 
9-33 
9-33 

4-31 

4-33 
4-36 
4.36 
6.26 
6.47 

6.59 
6.79 
6.78 
6.78 

7-03 

7-45 
7.63 
7-63 
8.04 

9-53 

9-51 
9.64 
9.64 

10.21 
10.31 

10.31 

10.37 

3.22 

3-24 

3.26 
3.26 
4-69 
4-85 

4-93 
5.o8 
5.o8 

s!?6 

5.58 
5.71 
5-71 

6.02 

7.T5 

7.12 
7.22 

7!66 
7-73 

7.73 
7.77 
8.53 
8.59 
9.09 

2.47 

2.48 
2.50 
2.50 
3.6o 
3-72 

3.79 
3-9° 
3-9° 
3-9° 
4.04 

4.28 
4-37 
4-37 
4.62 

5.49 

5.45 

5-55 

I'M 
5.94 

5.94 
5.97 

6.55 
6.59 
6.98 

Paterson  
Trenton  
Phoenix 

Pottsville 

Trenton  
Pittsburgh  
Buffalo  

Pottsville  

Trenton  
Pittsburgh  
Buffalo 

Paterson  

Trenton     

Pottsville  

Trenton  

Pittsburgh  
Pottsville  
Phoenix  
Buffalo      

V 

•• 

529 


TABLE    XIX.— ( 

ROLLED-IRON   BEAMS   IN   FIRST-CLASS    STORES. 

DISTANCES  FROM  CENTRES  (in  fcef). 
See  Arts.  694,  697,  6O9  and  7OO. 


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529  a 


TABLE    XIX.— (Continued.} 

ROLLED-IRON    BEAMS   IN   FIRST-CLASS   STORES. 

DISTANCES  FROM  CENTRES  (in  feet}. 
.See  Arts,  694,  697,  699  and  7OO. 


LENGTH  (in  feet)  BETWEEN  BEARINGS. 

0 

M 

S.    g  oiS  SF2     fev5S8.K 

fc«sssa  vSvSsa  " 

H            MWONO            NN(NNC1            NCOfOCOm          m  CT  f*>  CO 

0 

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SK    ^3«ffa£?    ^?KS^     ^,?KK^     ^5-^    ^3tf# 

£1 

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529  b 


TABLE   XX. 

See  Arts.   7OI,    7O5  and  7O6. 


The  larger  figures  five  the 
average^   for    use    in    the 
rules. 

See 
Table 
XLIL 

For- 
mula 
(10.) 

It 

See 

Table 
XLllI. 

For- 
mula 

IIS 

II 

fe. 

Formula 
(117.) 

11 

<a 

For- 
mula 

"ft 

11 

St* 

Table 
XLIY. 

D  OJJ^ 
2 

hi) 

See 

Table 
XLV. 

i  0~(i 

D  '£.  W 

HC/3     .     . 

See 

Table 
XLVI. 

Z  H 

*& 

gg«*n 

2o'i8< 

11>1 

e^ 

«  ' 

& 

Georgia  Pine  •< 

850 
1176 
952 
I2OO 
1406 
460 
650 
875 
417 
550 
722 

420 
500 
643 

280 
450 
707 
580 
600 
700 
442 
480 
520 
860 
9OO 
960 
1067 

IIOO 

1167 
1040 

1050 

IIOO 

616 
650 
746 
725 
750 
824 

507 
650 

790 

813 

850 
920 

394 

557 
490 
460 

475 
589 

4807 
5900 
6990 
4470 

5050 
5650 
1704 
3100 

4444 
2209 
3500 
4819 
2026 
2900 
3766 
1660 
2800 
4000 

3450 
3556 
2300 
2550 
2824 
3800 
4000 
4248 
4962 
5150 
5333 
37°4 
3850 
4000 
2800 
2850 
2933 
3619 
3900 
4211 

3273 
3600 
3894 
4545 
4750 
5000 

2022 
3350 
2697 
2249 
2580 
4466 

•001069 
•OOIO9 

•OOII12 
•001239 
•OOI5 
•001764 
•000791 
•  00086 
•00093 
•0008646 
•OOOgS 
•0010987 
•OOIOI56 
•OOI4 
•001791 
•000937 
•00095 
•000971 
•0009375 
•  OOO96 
•0009896 
0008854 
•OOIO3 
OOII77 
•OOIO42 

•OOIII 

•001177 
•0013854 
•0014 

•0014063 
•001198 

•  0013 

•001406 
001563 

•001563 
•001563 

•00099 
•OOIO4 
•001094 
•001146 
•OOII6 
•001177 
•001042 
•OOIOg 
•001146 
•0009014 
•0007256 
•0009014 
•OOO8IO4 
•000834 
•OOO6I2 

i-35i4 
1-8357 
2-1013 

2-3874 
2  •  2002 
1-9593 
4-7400 

2-9405 

3'°324 
2-227I 
i  •  8940 
2-8350 
I  -7105 
i   3240 
2.5002 
2  •  3496 

2    5282 
2    5512 
2'5l6l 
2-7628 
3    0146 
2--5382 
2-I728 
3-OI66 

2-8153 
2    6667 

2-I558 
2-IIgO 
2-l6l2 

3-2552 
2-9I38 
2    7165 

1-9549 
2-0266 

2    2601 
2-8I05 
2-5682 
2    4842 

1-8773 

2-1618 
2-3940 
2-3843 

2  •  2SO2 
2-2300 
3-0024 
3-I826 

2-7994 

3-5054 
3-0660 
2-9930 

11671 
16000 
21742 
11487 
24800 
33882 

"453 
19500 

I57I9 
19500 
22069 

I200O 
9871 

7^3 
840 

934 
970 
Il6o 
1389 
1076 
1250 
1474 
463 
540 
647 

433 
480 
530 
322 
370 
410 

8i7o 

9500   ; 
JI5<>3   i 
1  1009 
11700 
12582 

6531 
8000 

9775 
7166 

7850  ! 
8408 

58-9   i 
6650 
7502   ! 

5213 
5-00 

6281 

White  Oak               .  .    .  -! 

Spruce                            .  .  -< 

White   Pine                     \ 

Hemlock  \ 

Whitewood                       -) 

Chestnut  \ 

Ash                 .                    -j 

Maple                                \ 

Hickory  ....        .            \ 

Cherry              .                 J 

Black  Walnut  -J 
Mahogany.  St.  Dom  .  .  J 

Bay  Wood.  J 
Oak,  English  

"       Dantzic.  .  .  . 

Adriatic  

1  Oaks,  Average  of 

I  Oak,  Canadian  

TABLE    XX.— (Continued.} 


See  Arts.    7O  1 ,    7O3  and  7O6. 


The   larger  figures  gh>e  the 
aver  -age  \    for    use    in    the 
rules. 

See 

Table 
XL1I. 

For- 
mula 
(10.) 

li 
CQ 

See 
Table 
XLIII. 

For- 
nmla 

(113.) 

£ff 

M* 

ii 

s 

Formula 
(117.) 

n 

1! 

* 

For- 
mula 
(118.) 

iR 

t>. 

II 

i 

See 
Table 
XLIV. 

0,  Z  U  t>< 

D  O  Id 
tt  *«>   II 
2 
0  W  I  «T 

hH  H  u 

jt«l 

^M6-j 

f    M     ""'     < 

^"«§ 

sssip 

See 
Table 
XLV. 

:-di 

«I! 

0   ^C/J 
HC/3     .,c 

8^S^ 

r?-- 

fe«   < 

-  !  «  of  n 
agS  c 

RESISTANCE  TO  ^ 
CRUSHING,  PER  SQ.  IN.  ^^^ 
SEC.  AREA,  SHORT  \^  | 
BLOCKS.  =  C. 

^ 

e< 

Ash   

675 
519 
338 

544 
447 
367 
369 
350 
360 
38i 
421 
408 
284 
277 
376 
330 
491 

2OOO 
2500 
3OOO 
l6oO 

2100 
2600 
2400 
26OO 
2800 
1600 
19OO 

22OO 

3200 
6OOO 
72OO 

3810 
3134 
1590 
2836 

4259 

3454 
3080 
2293 
2686 
1858 
2013 
1961 
1422 
2078 

2437 
2093 

3375 
41500 
50000 
58500 
27700 
40000 

532co 
55500 
62OOO 
69000 
53000 
6OOOO 
67000 
60000 
65000 
71500 
67000 
7OOOO 
74000 

•0007177 
•000582 
•0009552 
•0006428 
•0004279 
•0005277 
•0004932 
•0006813 
•0005868 
•  cooS  i  74 
•0007762 
•  0007903 
•0010686 
•0006265 
•0006412 
•000744 
•0006173 

3-4285 
3.9520 
3-0910 
4-1446 
3-4066 

2  •  7966 
3-3738 
3-III7 
3-I723 
3-4843 
3-7423 
3-6564 
2-5958 
2-9552 
3.3420 

2-9433 
3-2732 

2OOOO 
27000 
45000 
T3000 
17OOO 
26000 
40000 
6OOOO 
80000 
3OOOO 
50000 
65000 

1I5780 
155500 
190262 

80000 
I2OOOO 
170000 
Socoo 
IOOOOO 
140000 
40000 
70000 
looooo 
40000 
5OOOO 
65000 

IOOO 

3000 

6000 

IOOO       1 

2000 
3000 
4000    j 
20000 

Beach 

Elm                  

Pitch  Pine  

Red   Pine                       .    . 

Fir    New  England     .  .  . 

41      Ricra     

"      Average  of  Riga  .  . 
*'      Mar  Forest          .  •  . 

Av'ge  of  Mar  Forest 
Larch 

« 

Average  of.  
Norway   Spar  

Cast-Iron,  American..  J 
English..  ...j 
"Wrought-I  ron,  Amer  .  3 
English  J 
"-              Swedish  3 
Steel  Bars                         J 

.0002 



"      Chrome  •< 

Blue  Stone   Flagging  J 
Sandstone  -j 

Brick,  Common.  3 
*'        Pressed  

122 

2OO 

251 

33 

59 
94 
20 

33 

43 

37 

147 

Marble,  Eastchester.  .  .  . 

' 

531 


TABLE  XXI. 

SOLID    TIMBER     FLOORS. 

DEPTH  OF  BEAM  (in  inches). 

See  Art.  7O2. 


LENGTH  BETWEEN  1 
BEARINGS  (in  feet).  || 

DWELLINGS  AND  ASSEMBLY  ROOMS. 

FIRST-CLASS    STORES. 

i2     .                   td 
O  W                   U 

ei  Z                  £> 

H   W 

5  2 

HEMLOCK. 

GEORGIA 
PINE. 

SPRUCE. 

M 

t  w 

HEMLOCK. 

8 
9 
1O 

2-4O 
2-72 
3-03 

2-83 
3-20 

3-56 

3-oi 
3-40 
3-79 

3-04 
3-43 

3-82 

3-i5 

3-55 
3-95 

3-73         3-97 
4-20         4-47 
4-68          4.'.  8 

4-01 
4-52 
5-03 

11 

3-35 

3-93 

4-18 

4-21 

4-35 

5-15 

5-48          5-54 

12 

3-67 

4-30 

4-58 

4-61 

4-76 

5-63 

5-99     !     6'°5 

13 

3-99 

4-68 

4-98 

5-01 

5-16 

6-  10 

6-50 

6-56 

14 

4-32 

5-05 

1  5-38 

5-4i' 

5-57 

6-58 

7-01 

-•07 

15 

4-64 

5-43 

5-78 

5-Si 

5-98 

7  -06 

7-52 

7-59 

16 

4-97 

5-8i 

6-19 

6-21 

^•39 

7-55 

8-03          8-io 

17 
18 

5-^4 

6-19 

6-58 

6'59 
7-00 

6-62 
7-03 

6  -So 

7-21 

8-03 
8-51 

8-55          8-62 
9-06          9-14 

19 
20 

,5-98 
6-32 

6-97 

7-35 

7-83 

7-44 
7-85 

7-^3 

8-05 

9-00 
9-48 

9-5S 

10-  10 

9-66 
10-18 

21 

6-66 

7-75 

8-24 

8-27 

8-46 

9-97 

10-61 

10-70 

22 

7-00 

8-14 

S-66 

8-68 

S-S8 

10-46 

11-13 

11-22 

23 

7-35 

8-53 

9-08 

9-10 

9-30 

10-95 

n-66 

11-74 

24 

7-70 

8-93 

9-5i 

9-52 

9-72 

11-44 

12-18 

12-27 

25 

8-05 

9-33 

9'93 

9'94 

10-15 

11-94 

12-71 

12-79 

26 

8-40 

9-73 

10-36 

10-37 

10-57 

12-43 

13-23 

13-32 

27 

8-76 

10-14 

10-79 

10-79 

II  -OO 

12-92 

13-76 

I3-85 

23 

9-  II 

10-54 

11-22 

11-22 

II  -42 

13-42 

14-29 

14-38 

29 

9-47 

10-95 

II  -65 

II-65 

11-85 

13-92 

14-82 

14-91 

3O 

9-83 

1  1  -  36 

I2-Og 

12-08 

12-28 

14-42         15-35 

15-44 

532 


TABLE    XXII. 


MATERIALS     USED     IN     THE     CONSTRUCTION    OR    LOADING    OF 

BUILDINGS. 

WEIGHTS  PER  CUBIC  FOOT. 

As  per  Barlow,    Gallier,    Ilaswell,    Hursf,    Rankine,    Tredgold,    Wood 
and  the  Author. 


MATERIAL. 

I 
o 

K 

fe 

o 

H 

AVERAGE. 

MATERIAL. 

~ 
ta 

0 

AVERAGE. 

WOODS. 

41 

35 
49 
4i 

39 
35 
59 

27 

47 
3^ 

32 
29 
27 

27 

21 

69 

51 
5i 
51 
57 
53 
49 
65 

35 
57 

38 

46 
4i 
55 

4i 

33 
83 

46 
38 
50 
49 
46 
42 
62 
83 
64 
31 
52 
34 
40 
39 
35 
41 
15 
34 
4O 
44 

1? 
IS 

43 
46 
27 
37 
47 
53 
62 
37 
26 
49 
52 
33 
43 
23 
62 
46 
57 
38 

Mahogany,  St.  Domingo.  .  .  . 
Maple  

45 
33 
35 

60 

38 
57 
47 
43 

4° 
38 

63 

49 

55 

66 

79 
54 
57 

44 
S^ 

55 
41 
45 
62 
63 
54  ! 
47  ! 
54 
68  i 
51 
5O 
58 
44 
42 
48  ! 
43 
32  i 
37 
39  i 
28  : 
33 
45 
30 
44 
23 
45 
3O 

•n 

38 
51 
30 
8O 
33 
49 
27 
50 

614 
5O6 
544 
531 
516 

Mulberry   ...    
!Oak,  Adriatic  
"      Black   Bog  
"     Canadian  
Dantzic  
English  

"      Red."i 
'•      White  
Olive  
Orange    
Pear-tree.             
Pine,  Georgia  (pitch)   
Mar  Forest               .    . 

Alder 

Apple-tree  

Ash                             

Beech 

Birch.    ..    . 
Box  

"     French  
Brazil-wood  

Cedar  

"     Canadian  

"     Palestine.    
"     Virginia  Red         
Cherry    

Chestnut,  Horse  

11      Memel  and  Riga 
"      Red    .:.  ... 
''      Scotch   
"      White    
'*      Yellow...  
Plum   
Poplar.    .                         .... 

29 

27 

21 

27 
41 
23 

55 
24 
36 
4i 

g 

40 
25 

487 

528 
508 

35 

Si 
35 
39 
49 
37 

59 
36 

£ 

83 
4o 
58 
29 

525 

S34 
524 

Cork  
Cypress  

Spanish  
Deal,  Christiania  

''       English  

u       (Norway  Spruce).    ... 
Dogwood  
Ebony  

1   Quince  ..... 
Redwood 

Rosewo  >d 

Elder    

Sassafras  
Satin  wood  
Spruce      
Sycamore. 

Elm   
Fir  (Norway  3pru<-e)  . 

33 

21 
30 

59 
33 
44 

"    (Ued  Pine)  
"    Rig;l  

Teak  
Tulip-tree  
Vine.    
Walnut,   Black  
"•         White 

'•       Water 

3* 

58 
63 
35 

54 

83 
Si 

40 

'  Hackmatack  

21 
40 
41 
31 
31 

41 
41 

35 

Hemlock  

Hickory  

Whitewood   . 
Yew 

Larch  

METALS. 
Bismuth,  Cast  ,. 
brass,  Cast  
''        (Gun-metal)  .  .  , 
Plate  

"      Red. 

"      White  
Lignum-vitae    

Logwood  
Mahogany,  Honduras  

i          

Bronze    .  . 

533 


TABLE    XX II.— (Continued^ 

MATERIALS    USED     IN     THE    CONSTRUCTION     OR    LOADING    OF 

BUILDINGS. 

WEIGHTS  PER  CUBIC  FOOT. 

As  per  Barlo-w,    Gallier,    Haswell,    Hurst,    Rankine,    Tredgold,    Wood 
and  the  Author. 


MATERIAL. 

I 
h 

O 

H 

AVERAGE. 

MATERIAL. 

0 
K 
t* 

o 

C-i 

ui 
a 

< 

M 

H 

< 

1O4 
100 
112 

no 

130 

81 
10O 

113 
145 
122 
16O 
96 
83 
79 
85 
54 
13O 
106 
110 
126 
126 
250 
160 
185 
163 
16O 
183 
165 
163 
174 
165 
164 
166 
185 
166 
105 
134 
140 
52 
169 
146 
162 
170 
166 
170 

Copper,  Cast  

537 

549 

543 
556 
544 
1206 
1108 
509 
481 
454 
475 
480 
7O9 
717 
713 
851 
849 
837 
488 
453 
975 
1345 
1379 
142 
636 
655 
658 
644 
489 
462 
439 

173 
156 
8O 
277 
171 
139 
129 
160 
10-' 
138 
1O7 
100 
1O5 

Brick-work  ... 

96 

112 

"         Plate                   

... 

12O 

Gold  

"          in  Mortar  

100 

475 
434 

487 
474 

Iron,  Bar  
"      Cast  
"       Malleable 

Roman,   Cast  

and  Sand, 
equal  parts.. 
Chalk 

iie 

119 

'is 

77 

46 
125 

J74 

125 

102 
90 

81 

§6J 

'35 

125 

^65 
195 

"       Wrought  

474 

486 

Lead    Cast 

Clay  
''     with  Gravel  
Coal,  Anthracite  

English  Cast 

"       Milled       

... 

Bituminous  
"      Cannel  

"     60°  

Nickel,  Cast  

.. 

Coke 

Pewter..               

Concrete,  Cement  

Platina,  Crude 

k*         Pure 

F     t  h     C  '  ' 

95 

"     Rolled  
Plumbago    
Silver,  Parisian  Standard... 
Pure  Cast  
"           "        "     Hammered 
"       Standard  
Steel 

486 
456 
429 

165 

492 
468 
449 

180 

"         Loamy  

Emery  
Feldspar  
Flagging,  Silver  Gray     .    ... 
Flint  

155 
171 

Tin,  Cast 

"       Flint  

Zinc,  Cast  

STONES,  EARTHS,  ETC. 

Alabaster.  
Asphalt,  Gritted 

"       Plate 

153 
167 
158 

173 

181 
172 

"       White  
Granite  

"       Egyptian  Red  

Asphaltum  
Barytes,  Sulphate  of  
Basalt     
Bath  Stone 

57 
250 

155 

122 
124 

'85 

103 
304 

18£ 
156 

'34 
119 

"      Quincy  

Gravel 

90 

120 

Gypsum  

135 

'45 
199 

Beton  Coignet   
Blue  Stone,  Common  
Brick.  .  .  . 

Limestone  

»39 

Aubigne  

'     Fire-  

|    N.  R.  common  hard... 

... 

Marble 

161 

178 

'    Philadelphia  Front... 

... 

534 


TABLE    XX 1 1.— (Continued.) 

MATERIALS     USED     IN     THE     CONSTRUCTION    OR     LOADING    OF 

BUILDINGS. 

WEIGHTS  PER  CUBIC  FOOT. 

As  per    Barlow,    Gallier,    Hasivell,    Hurst,   Rankine,     Tredgoid,    Wood 
and  the  Author. 


MATERIAL. 

i 

0 

H 

AVERAGE. 

MATERIAL. 

o 
£ 

£ 

AVERAGE. 

Marble,  Eastchester  

167 

178 

173 
167 
166 
167 
I4O 
125 
175 
155 
98 
103 
107 

9 
86 
105 
1OO 

118 
83 
146 
72 
80 
18O 
175 
147 
56 
165 
165 
124 
112 
1O5 
105 
123 
105 
97 
172 
126 
144 
133 
142 
134 
141 
134 
162 
15O 

Serpentine. 

165 
144 
152 
95 
159 
167 
157 
18O 
135 
151 
14O 
160 
14O 
124 
115 
170 
76 

58 
49 
59 
62 
26 
20 
58 
57 
61 
17 
61 
69 
114 
73 
131 
559 
68 
171 
133 
131 
14 
8O 
62* 
64 
81 

Chester,  Pa  
''         Green 

... 

... 

"          Italian  
Marl                   

165 

100 

no 

169 
179 
140 

Slate       

137 

181 

150 
i€o 

Mica 

120 
1  2O 

'87 

88 

109 
118 

"     Welsh         

Mortar  

Stone,   Artificial  

Stone-work  
Hewn    

"           Rubble 

"        Hair,  incl.  Lath  and 
Nails,  per  foot  sup. 
Hair,  dry  
44       new...    
"    Sand  3  and  Lime  paste  2 

well  beat  together 

7 

ir 

Tiles,  Common  plain  . 
Trap  Rock  

... 

MISCELLANEOUS. 
Ashes.  Wood       .... 

Peat,  Hard  
Petrified  Wood  
Pitch  

... 

Plaster.  Cast  
Porphyry,  Green  
Red  
Portland  Stone  
Pumice-stone   .    .    
Puzzoiana  
Quartz,  Crystallized  
Rotten-stone    

132 

161 

Butter  
Camphor  
Charcoal  

17 
14 

52 

IO 
56 

34 
25 

'62 

24 

66 

Cotton,  baled  
Fat  

Gutta-percha  
Hay   baled                 

Sand,  Coarse    
Common  
•      Dry  
Moist  
4      Mortar.. 

92 
90 
118 

118 

120 
128 

101 

158 

... 

•      Pit  

92 

Quartz  

Red  Lead  

with  Gravel  

Resin          

Sandstone  
Amherst,  O.  . 
Belleville,  N.J... 
Berea,  O  
Dorchester,  N.  S. 
Little  Falls,  N.  J. 
44          Marietta,  O.       .  . 
**          Middletown,  Ct.. 

130 

Rock  Crystal... 

Salt  

20 

100 

Saltpetre.     .                   .. 

8 
60 

Water,  Rain  ...    . 

»»       Sea    

Whalebone  

535 


TABLE    XXIII. 

TRANSVERSE   STRAINS    IN    GEORGIA   PINE. 

LENGTH  i  •  6  FEET.  BETWEEN  BEARINGS. 

See  Arts.  7O4  and  7O5. 


NUMBER  OF    / 
EXPERIMENT.  ( 

1 

2 

3 

4 

5 

6 

7 

8 

9 

i 

DEPTH       I 

1 

(/«  inches).   \ 

1-04 

i  04 

1-03 

i  04    I     1-03 

j 

i  03 

1-03 

1-03 

I    02 

BREADTH     ( 
(in  -inches),   j 

i  05 

1-04 

i  03 

1-03    I    1-04 

1-04 

1-04 

1-03 

1-04 

PRESSURE 
(in  founds). 

DEFLECTION  (in  inches). 

0 

000 

ooo 

ooo 

•ooo 

ooo 

•ooo 

coo 

oco 

•000 

25 

020 

015 

020 

01  5 

015 

•OiD 

010 

020 

•015 

50 

•040 

030 

•040 

030 

•025 

•035 

•025 

•040 

•030 

75 

•OSS 

040 

060 

•04S 

045 

•050 

•035 

OSS 

•045 

100 

•070 

•050 

oSo 

•060 

•065 

•065 

-050 

•070 

•  -060 

0 

•ooo 

ooo 

000 

•ooo 

000 

•000 

•000 

oco 

•oco 

100 

•070 

050 

•080 

060 

c65 

•065 

•050 

•070 

•ceo 

125 

•085 

065 

095 

•°75 

•080 

•080 

•070 

•090 

075 

i5O 

•TOO 

080 

•  1  10 

•090 

•  100 

•090 

085 

•no 

085 

175 

•"5 

•095 

-125 

105 

115 

•  1  05 

•  ICO 

130 

•100 

200 

•130 

no 

•140 

•120 

•  130 

•  1  20 

115 

•15,0 

•115 

0 

000 

•ooo 

•ooo 

•000 

ooo 

•ooo 

•ooo 

•000 

coo 

200 

•  130 

•  no 

•  140 

•  1  20 

130 

120 

•i  '5 

•  150 

•  115 

225 

•145 

•120 

•160 

•'35 

•145 

•'35 

•125 

•  170 

130 

250 

•160 

•135 

•175 

ISO 

160 

•150 

•140 

i$p 

-J45 

275 

-I75 

•150 

•  IQO 

•<6s 

•180 

•ico 

160 

•2IO 

•160 

300 

•190 

•160 

-210 

•180 

•IC;5 

•»7i 

•*T5 

225 

175       | 

O 

•ooo 

•ooo 

•OOO 

•ooo 

•ooo 

ooo 

•ooo 

•ooo 

•coo     1 

3OO 

•IQO 

160 

210 

180 

•!95 

•175 

•175 

•22<y 

•175 

325 

•2IO 

•'75 

•230 

200 

•215 

•190 

•190 

•?45      j        IQO 

350 

•225 

•IQO 

•2^O 

•  215 

•230 

•2IO 

•2IO 

•26^ 

205 

375 

•240 

•205 

•265 

•230 

•245 

•225 

•225 

•285 

220 

Eoo 

•255 

•  220 

•280 

245 

•260 

•240 

•245 

•310 

235 

o 

oco 

CCO 

005 

coo 

•ooo 

•005 

•005                -CO5 

co^ 

4OO 

•255 

•22O 

•2£o 

•245 

•260 

•240 

•245               -310 

240 

425 

•275 

-235 

•300 

265 

•  280 

•255 

•255 

•330  ;   -255 

45O 

•290 

•250 

•320 

•280 

295 

•270 

•270 

•355           -275 

475 

•310 

•265 

•340 

•295 

•'3  '5 

285 

•28; 

385           -2co 

5OO 

•330 

•280 

•380 

•3J5 

•335 

•300 

•3CO 

•4'5           "SOS 

0 

•ooo 

•OIO 

•030 

•000 

•005 

•005 

•005 

•030 

•cos 

50O 

•330 

•280 

•380 

•3'S 

•340 

•300 

•300 

•430 

•305 

525 

•35° 

•  295 

•330 

•360 

'3'5 

•  32O 

-470 

•325 

550 

•370 

-3'0 

•350 

•380 

•330 

•33* 

•345 

575 

•390 

•330 

•375 

405 

-350 

•355 

•370 

GOO 

4i5 

•35° 

•395 

435 

"375 

•375 

•400 

0 

•020 

•02O 

•015 

•025 

•020 

•025 

•025 

600 

•420 

•350 

.... 

•4°5 

.440 

•380 

•380 

•410 

625 

•460 

•370 

•435 

•475 

•405 

•405 

•455 

65O 

•560 

•400 

•470 

•535 

•430 

•43° 

675 

•440 

•460 

•475 

700 

•480 

BREAKING 

WEIGHT  (In 

674- 

719- 

5l8- 

6C2- 

652- 

699- 

685- 

536- 

642- 

pounds). 

536 


TABLE  XXIV. 

TRANSVERSE  STRAINS  IN  LOCUST. 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.    7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-   V 

1O 

11 

12 

13 

14 

15 

16 

17 

18 

19 

20 

MENT.     \ 

DEPTH   | 
(in  inches),  f 

1-03 

1-08 

1-05 

I  -08 

1-07 

1-65 

1-07 

I  -08 

I  -07 

1-09 

I  -08 

BREADTH  j 
(in  inches).  \ 

I  -02 

1-05 

I  -08 

I  -02 

1-09 

I  -08 

1-68 

1-04 

i-oS 

I  -08 

1-03 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches'). 

0 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•660 

•OOO 

•OOO 

•OOO 

•660 

25 

•015 

•015 

•015 

•  015 

•010  -015  -015 

•615 

•015  -615  -015 

5O 

•035 

•030 

•030 

•030 

•625 

•0301  -030  ,  -035 

•030  -030  -030 

75 

•055 

•050 

•045 

•045 

•635  '656:  -045 

•656 

•645 

•645  -645 

100 

•075 

•065 

•060 

•060 

•O5O   -065  ;  -O6O 

•676 

•660 

•666  -060 

0 

•000 

•000 

•000 

•OOO 

•  OOO   •  OOO   •  OOO 

•66O 

•000 

•OOO  -OOO 

100 

•075 

•065 

•  060  j  •  060 

•656 

•065  :  -O6O 

•670 

•  060 

•060 

•e66 

125 

•095 

•080 

•075  1  -075  -065 

•080  1  -075 

•QgO 

•075 

•075  -080 

15O 

•no 

•100 

•090 

•  090  •  080 

•095  -090 

no 

•090 

•090  -095 

175 

•125  -115 

•105 

•110 

696 

•115  -105 

•125 

•165 

•IOO  -IIO 

200 

•145 

•  130 

•115 

•125 

•165 

•130  -120 

145 

•120 

•115 

•125 

O 

•OOO 

•OOO 

•OOO 

•000 

•OOO 

•  OOO   •  OOO 

•OOO 

•000 

•000 

•666 

200 

•145 

•  130 

•115 

•  125 

•165 

•130;  -120 

•145 

•120 

•  115 

•125 

225 

•165 

•  150 

•136 

•  140 

•126 

•145 

•135 

•165 

•136 

•  130 

•  1401 

25O 

•180 

•  170 

•145 

•155 

•135 

•  165 

•156 

•  190 

•145 

•145 

•155 

275 

•200 

•185 

•160 

•175 

•145 

•180 

•I65 

•2IO 

•160 

•160 

•  170 

3OO 

•22O 

•200 

•175  -19° 

•166 

•195  -180 

•230 

•175 

•175 

•190 

O 

•OOO 

•005 

•OOO   -OOO 

•600 

•  OOO   •  OOO 

•OIO 

•ooe 

•000 

•  005 

300 

•226 

•205 

•175  -19° 

•  160 

•195  -180 

•230 

•175 

•175 

•190 

325 

•240 

•220 

•  190  -2IO 

•175 

•210  -195 

•250 

•  190 

•185 

•205 

350 

•26O 

•240  -205   -225 

•  190 

•230  -2IO 

•  270 

•205 

•200 

•22O 

375 

•275 

•265 

•22O   -240 

•205 

•245   -22O 

•290 

•22O 

•215 

•  240 

40O 

•295 

•290 

•235  -255 

•22O 

•265   -235 

•315 

•235 

•  230 

•255  1 

0 

•005 

•02O 

•005  -005 

•OOO 

•005  !  -005 

•OOO 

•665 

•OIO  | 

4OO 

•295 

•295 

•235  -255 

•22O 

•265 

•235  : 

.... 

•235 

•236 

•255: 

425 

•335 

•250 

•275 

•235' 

•280 

•  250; 

.... 

•250 

•245 

•275 

45O 

.... 

•265 

•295 

•250 

•295 

•265 

.... 

•265 

•  266 

•  290  ;  j 

475 

•280 

,  o  rr\ 

*  °f)  " 

•  280 

5OO 

•  290 

•325 

•280 

•335 

•295 

.... 

•295 

•275 
•290 

•3IQ  ; 

•336  ; 

537 


TABLE    XXIV.— (Continued.} 


TRANSVERSE     STRAINS    IN     LOCUST 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF  ) 

i   EXPERI-   V 
MENT.    ) 

1O 

11 

12 

13 

14 

15   16 

1 

17 

18 

19 

20 

DEPTH  \ 
'  (in  inches),  f 

BREADTH  ( 
(in  inches).  \ 

1-03 
I  -O2 

I  -08 
1-05 

1-05 
i  -08 

i  -08 
i  -02 

1-07 
1-09 

1-05 
i  -08 

1-07 
i  -08 

i  -08  j  1-07  \  1-09 

; 
I  •  04  I  I  •  08   I  •  O8 

I  -08 
I  -03 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

0 
500 
525 
550 
575 
600 

O 
600 
625 
65O 
675 
7OO 

0 
700 
725 
750 
775 
800 

0 
SOO 
825 
850 
875 
90O 

O 
900 
925 
950 
975 
1000 

•005 
•290 
•305 
•320 
•335 
•350 

OIO 

•350 
•365 
•385 
•400 

•4i5 

•015 
•4i5 
•435 
•455 
•475 
•495 

•035 
•505 
530 
•555 
•585 
•615 

065 
•635 

.66e 

•005 
•325 
•345 
•365 
•385 
•405 

•020 

•405 
•430 

•455 
•485 
510 

•035 
•515 
•540 

•575, 
•610 
640 

•075 
•650 
-690 

•725 

•005 
•280 

•295 
•310 

•325 
•345 

•OIO 

•345 
•365 
•385 
•405 
•425 

•030 
•430 

•455 
•475 
•505 
•535 

•060 

•545 
•580 
•615 
•645 
•675 

015 
•335 
•355 
•375 
•395 
•415 

020 
•420 

•455 
•49° 

•52< 
•560 

•040 

•565 
•595 
•635 

•005 

'-95 
•310 

•325 
•340 
-360 

015 

.360 

•375 
•390 
•405 
•420 

•020 
•420 
•440 
•460 

•480 
•500 

•035 
•505 
•530 
•550 

•575 
•605 

•050 
•610 
•640 
•670 
•700 
•735 

.... 

•005 
•295 
•310 

•325 
•340 

•355 

•OIO 

•355 
•375 
•39° 
•405 
•425 

•025 

•425 
•445 

•465 
•490 

•515 

•045 
•520 

•545 
-570 
•600 
•630 

•075 
•640 

•675 

•005 
•290 
•305 
•320 

•335 
•350 

-015 
•355 
•370 
•390 
•410 
•430 

•025 
•435 
•455 
•480 
•500 
•520 

•045 
•530 
•560 

•585 
•615 
•650 

•015 
•330 
•350 
•3/0 
•39° 
•4i5 

•025 
•4i5 
•440 
•460 

•485 
•510 

•050 
•520 

.... 

.... 

.... 





.... 

.... 





.... 

• 

•695 

•  7^ 

i  BREAKING  j 
I  WEIGHT  (in  > 
pounds).   ) 

| 

449-  425-  1046-  956-  looi-  860- 

1037- 

402-   937'  ;I027- 

7i5- 

TABLE    XXV. 

TRANSVERSE    STRAINS    IN   WHITE   OAK. 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.    7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-  y 

21 

22 

23 

24 

25 

26 

27 

28 

29 

3O 

MENT.    ) 

DEPTH   j 
(in  inches),  j 

I  -06 

I  -06 

I  -08 

1-07  i  -08 

I  -06 

,.„ 

i  -08 

1-07 

1-06 

BREADTH 
(in  inches). 

I  -08 

i  -08 

I  -06 

i  •  06  i  •  06 

i  -08 

1-07 

1-07 

I  -06 

i  -08 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches'). 

0 

•  ooo   •  ooo 

•ooo 

•ooo 

•ooo 

•ooo 

•000 

•ooo 

•ooo 

•ooo 

25 

•030   -040 

•035 

•040 

•045 

•045 

•035 

•040 

•035 

•045 

5O 

•060   -075 

•065 

•080 

•090 

•085 

•075 

•080 

•070 

•085 

75 

•  090   -115 

•095 

•115 

•135 

•  125 

•115 

•125 

•  105 

•125 

10O 

•120    -155 

•  130 

•150 

•180 

•  170 

•155 

•165 

•  140 

-165 

0 

•  ooo   •  ooo 

•ooo 

•000 

•ooo  -GOO 

•000 

•000 

•ooo 

•ooo 

1OO 

•  1  20 

•155 

•130  i  -150 

•185  -170 

•155 

•  165 

•  140 

•  170 

125 

•  150 

•195 

•  165  •  190 

•235 

•215 

•200 

•  215 

•175 

•215 

150 

•180  |  -235 

•195  -230 

•295 

•260 

•245 

•  260 

•2IO 

•260 

175 

•215 

•280 

•225  -275 

•355 

•  310 

•285 

•  310 

•240 

•  310! 

200 

•245 

•325 

•265 

•315 

•410 

•355 

•330 

1-360 

•275 

•365 

0 

•ooo 

•020 

•OIO 

•O2O 

•025 

•020 

•015 

•O2O 

•OIO 

•  025 

200 

•250  1  -330 

•265 

•325 

•410 

•365 

•345 

•360 

•  280 

•365 

225 

•  285  -380 

•  310 

•380 

•480 

•420 

•395 

•420 

•320 

•420 

250 

•320  -440 

•350  -430  -555 

•480 

•450 

•485 

•360 

•  480  1 

275 

•360  -500 

•390  -480 

•<>35 

•545 

•SIS 

•560 

•405 

•545 

30O 

•400 

•560 

•440  -540 

•715 

•615 

•580 

•640 

•450 

•615 

0 

•  030 

•060 

•040 

•065 

•  100 

•075 

•065 

•080 

•045 

•080 

300 

•415 

•530 

•440 

•560 

•735 

•635 

'595 

•660 

•455 

•640 

325 

•465 

•650   .490   -620 



•705 

•665 

•510 

•725  • 

350 

•515 

•7J5 

•545   -680 

•*6s 

375 

•570 

•605 

•  760 

.... 





•62S 

*4OO 

•  630 

•670 

•600 

0 

•  105 

•  ion 

400 

•690 

.  710 

425 

— 



•760 

BREAKING  ) 
WEIGHT  (in  > 
pounds).  ) 

520- 

404- 

510- 

475' 

368- 

430- 

426- 

391- 

504- 

40!' 

539 


TABLE   XXVI. 


TRANSVERSE     STRAINS     IN     SPRUCE. 

i 
LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.  7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-  v 

MENT.    J 

31 

32 

33 

34 

35 

36 

37 

38 

39 

40 

DEPTH  \ 
(in  inches).  ) 

1-09   1-05 

I  -08 

1-04 

I  -07 

1-04 

1-03 

1-07 

I  -08 

I  -04 

BREADTH  / 
(in  inches),  f 

1-04 

I  -08 

1-03 

1-07 

1-04 

I  -08 

1-07 

1-04 

1-04 

I  -08 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

O 

•  ooo 

•000 

•ooo 

•000 

•ooo 

•ooo 

•ooo 

•ooo 

•000 

•ooo 

25 

•020 

•025 

•040 

•025 

•025 

•030 

•025 

•025 

•025 

•02O 

50 

•040 

•045 

•075 

•050 

•050 

•060 

•045 

•045 

•045 

•040 

75 

•060 

•070 

•  no 

•070 

•075 

•090 

•065 

•070 

•065 

•060 

1OO 

•080 

•090 

•145 

•095 

•  ICO 

•  I2O 

•085 

•090 

•085 

•080 

0 

•OOO 

•ooo 

•000 

•ooo 

•000 

•OOO 

•ooo 

•ooo 

•ooo 

•OOO 

100 

•  080 

•090 

•145 

•095 

•  IOO 

•120 

•085 

•090 

•085 

•080 

125 

•  IOO 

•"5 

•180 

•115 

•  125 

•145 

•  no 

•115 

•  105 

•  IOO 

150 

•125 

•135 

•215 

•135 

•  150 

•175 

•135 

•  140 

•  125 

•  125 

175 

•145 

•155 

•250 

•155 

•  170 

•200 

•155 

•160 

•145 

•145 

200 

•165 

•175 

•285 

•175 

•  190 

•230 

•175 

.180 

•  165 

•165 

0 

•ooo 

•000 

•010 

•ooo 

•OIO 

•OO5 

•005 

•ooo 

•ooo 

•  005 

200 

•165 

•175 

•285 

•180 

•  190 

•230 

•175 

•  I  80 

•  170 

•165 

225 

•185 

•2OO 

•325 

•200 

•215 

•265 

•195 

•2OO 

•190 

•  190 

250 

•2IO 

•225 

3/o 

•220 

•240 

•295 

•220 

•225 

•210 

•2IO 

275 

•230 

•245 

•415 

•245 

•260 

•330 

•240 

•250 

•235 

•235 

30O 

•250 

•265 

•465 

•27O 

•285 

•370 

•260 

•270 

•255 

•255 

O 

•005 

•005 

•045 

•005 

•OIO 

•025 

•OO5 

•005 

•005 

•005 

30O 

•250 

•270 

475 

•275 

•285 

•375 

•260 

•275 

•255 

•255 

325 

•275 

•295 

•530 

•  300 

•310 

•410 

•285 

•300 

•275 

•280  1 

35.O 

•300 

•320 

-600 

•330 

•335 

•450 

•310 

•325 

•300 

•305 

375 

•330 

•350 

680 

•355 

•360 

•495 

•335 

•355 

•325 

•330 

4OO 

•380 

•390 

•760 

•385 

•390 

•540 

•365 

•395 

•350 

•360 

1 

O 

•035 

•030 

.... 

•020 

•020 

•070 

•020 

•025 

•OIO 

•O2O 

4OO 

•390 

•4OO 

.... 

•395 

•395 

•570 

•370 

•410 

•355 

•370 

425 

•460 

•440 

.... 

•425 

•  430 

•620 

•400 

•455 

•385 

•400 

45O 

.... 

•495 

•  ... 

•455 

•460 

•680 

•440 

•445 

•  440 

475 

.... 

•505 

.... 

•  485 

.... 

•500 

5OO 

.rfir 

•  ^7O 

O 

:>u:> 

j  /u 
•O7O 

500 

•59O 

•J? 

BREAKING 

WEIGHT  (in 

445' 

487- 

400- 

502- 

470- 

465- 

498- 

441- 

475' 

527- 

Pounds). 

540 


TABLE    XXVII. 


TRANSVERSE     STRAINS    IN     SPRUCE. 

LENGTH  i  •  6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-  > 

41 

42 

43 

44 

45 

46 

47 

48 

49 

50 

MENT.    ) 

DEPTH   I 
(in  inches).  \ 

1-52 

i-55 

1-56 

1-56 

1-56 

i-55 

i-55 

1-56 

i-55 

i'57 

BREADTH  I 
(in  inches).  ] 

1-09 

I  •  IO 

i  -06 

I  •  IO 

I  •  IO 

i  09 

1-08 

I-IO 

I-  IO 

1-09 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

O 

•OOO 

•OOO 

•OOO 

•000 

1 

•OOO 

•OOO 

•000 

•OOO 

•OOO 

•OOO 

25 

•OIO 

•OIO 

•OIO 

•OIO 

•OIO 

•OIO 

•OIO 

•OIO 

•005 

•005 

50 

•O2O 

•O2O 

•O2O 

•015 

•O2O 

•O2O 

•O2O 

•015 

•OIO 

•OIO 

75 

•025 

•030 

•025 

•025 

•030 

•  025 

•030 

•  025 

•O2O 

•O2O 

100 

•035 

•040 

•035 

•035 

•035 

•035 

•035 

•030 

•  025 

•025 

0 

•OOO 

•OOO 

•OOO 

•OOO 

OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

100 

•035 

•040 

•035 

•035 

•035 

•035 

•035 

•  030 

•025 

•  025 

125 

•045 

•050 

•045 

•040 

•045 

•045 

•040 

•035 

•030 

•035 

ISO 

•050 

•060 

•055 

•050 

•055 

•050 

•050 

•040 

•040 

•040 

175 

•060 

•070 

•065 

•060 

•065 

•060 

•055 

•  050 

•045 

•045 

200 

•065 

•075 

•070 

•065 

•070 

•065 

•065 

•055 

•055 

•055 

0 

•000 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•000 

•000 

•OOO 

200 

•065 

•075 

•070 

•065 

•070 

•065 

•065  -055 

•055 

•055 

225 

•070  -085 

•080 

•075 

•  080 

•070 

•070  -065 

•060 

•060 

250 

•080  -095 

•090 

•085 

•  090 

•  080  •  080  •  070 

•065 

•065 

275 

•090 

•IOO 

•IOO 

•090 

•095 

•085   -090  -075 

•075 

•075 

300 

•  IOO 

•no 

•  no 

•095 

•  105 

•  090 

•095 

•  085 

•080 

•  080 

0 

•OOO 

•OOO 

•000 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

•OOO 

.  300 

•IOO 

•  no 

•  no 

•095 

•  105 

•  090 

•095 

•  085 

•  080 

•  080 

325 

•  105 

•US 

•120 

•  105 

•115 

•  IOO 

•IOO 

•090 

•085 

•  085 

35O 

•«5 

•  I2O 

•130 

•115 

•  125 

•  105 

•  IIO 

•IOO 

•090 

•095 

375 

•120 

•130 

•140 

•125 

•130 

•110 

•120 

•  105 

•IOO 

•IOO 

400 

•125 

•140 

•150 

•135 

•  140 

•I  20 

•  125 

•IIO 

•105 

•IIO 

0 

•OOO 

•OOO 

•005 

•000 

•000 

•OOO 

•000 

•OOO 

•OOO 

•000 

400 

•  125 

•  140 

•150 

•135 

•140 

•I2O 

•J25 

•IIO 

•105 

•IIO 

425 

•135 

•  150 

•160 

•  140 

•145 

•  125 

•130 

•120 

•IIO 

•I2O 

450 

•145 

•160 

•170 

•145 

•155 

•135 

•140 

•  125 

•120   -125 

475 

•  150 

•165 

•180 

•155 

•165 

•140 

•150 

•  130 

•125 

•  130 

500 

•160 

•175 

•  190 

•  165 

•175 

•150 

•155 

•  140 

•I3o 

•135 

O 

•005 

•OOO 

•OIO 

•OOO 

•000 

•OOO 

•000 

•OOO 

•OOO 

•OOO 

500 

•  160 

•175 

•  190 

•165 

•175 

•  150 

•155 

•  140 

•  130 

•135 

525 

•  170 

•180 

•  205 

•  170 

•185 

•155 

•160 

•145 

•  140 

•145 

550 

•175 

•190 

•215 

•180 

•195 

•160 

•170 

•  150 

•145 

•155 

575 

•185 

•200 

•  225 

•  185 

•  205 

•  170 

•180 

•  160 

•  150 

•160 

600 

•190 

•2IO 

•;>40 

•195 

•  215 

•180 

•185 

•  170 

:i6o 

•  170 

541 


TABLE    XXVIL— (Continued) 


TRANSVERSE     STRAINS     IN     SPRUCE 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

'See  Arts.    7O4  and  7O5. 


NUMBER  OF  ) 

i 

EXPERI-  >• 

41 

42 

43 

44 

45 

46 

47 

48 

49 

50 

MENT.    ) 

DEPTH   | 
(in  inches),  j 

1-52 

i-55 

1-56 

1-56 

1-56 

i'55 

i-55 

1-56 

i-55 

i-57 

BREADTH  ) 
(in  inches).  \ 

1-09 

1  •  1O 

I  -06 

]  -10 

1-10 

i  -09 

i  -08 

1  -10 

I  -10 

i  -09 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

I 

0 

•005 

•005 

•015 

•005 

•005 

•005 

•005 

•005 

•ooo 

•ooo 

600 

•IQO 

•2IO 

•240 

•195 

•215 

•180 

•185 

•170 

•160 

165 

625 

•195 

•215 

•250 

•205 

•225 

•185 

•195 

•175 

•170 

•175 

650 

•205 

•225 

•265 

•215 

•235 

195 

•200 

•185 

•175 

•I  80 

675 

•215 

•230 

•275 

•220 

•245 

•200 

•2IO 

•  190 

•180 

•  190 

700 

•225 

•240 

•290 

•230 

•255 

•2IO 

•220 

•200 

•190 

•195 

0 

•005 

•005 

•025 

•OIO 

•OIO 

•005 

•005 

•005 

•005 

•ooo 

700 

•225 

•240 

•290 

•230 

•255 

•210 

•22O 

•200 

•  190 

•195 

725 

•235 

•250 

•305 

•240 

•265 

•22O 

•230 

•2IO 

•200 

•  205 

750 

•245 

•260 

•320 

•245 

•275 

•230 

-240 

•220 

•2IO 

•215 

775 

•255 

•265 

•335 

•255 

•  290 

•240 

•25O 

•230 

•22O 

•225 

8OO 

•265 

•275 

•350 

•  265 

•305 

•250 

•255 

•240 

•230 

•235 

0 

•OIO 

•OIO 

•040 

•OIO 

•025 

•OIO 

•010  -015 

•OIO 

•005 

80O 

•265 

•275 

•350 

•275' 

•  310 

•250 

•255  -240 

•23O 

•235 

825 

•275 

•285 

•370 

•285 

•330 

•260 

•265 

•250 

•240 

•245 

850 

•285 

•295 

•335 

•295 

•345 

•  270 

•275 

•260 

•255 

•255 

875 

.300 

•305 

•405 

•  310 

•365 

•  280 

•285 

•275 

•27O 

•265 

900 

•  310 

•315 

•420 

.320 

•405 

•  290 

295 

•290 

•290 

•275 

0 

•O2O 

•O2O 

•060 

•025 

•025 

•02O 

•030 

•O25 

•  015 

900 

•315 

•320 

•430 

•325 

.... 

•295 

•295 

•295 

•295. 

•275 

925 

•365 

•330 

•460 

•340 

•  310 

•310 

•310 

•315 

•290 

950 

•345 

.... 

•355 

.... 

-325 

•320 

•325 

•340 

•305 

975 

.... 

•360 

.... 

•370 

•340 

•335 

•345 

•470 

•320 

1000 



•370 

•335 

.... 

•365 

•350 

•365 



•335 

0 

•030 

•045 

•  050 

•030 

•060 

•035 

1OOO 

.... 

•380 

•395 

.... 

•375 

•355 

•400 

•345 

1025 

.... 

•390 

.... 

•400 

•375 

•450 

•365 

1O50 

.... 

•405 

.... 

.... 

•430 

•400 

•385 

1075 

•  415 

BREAKING 

! 

WEIGHT  (in 
pounds). 

950- 

1074- 

926-  !  ioo  i  • 

900- 

1067-  1071- 

1028- 

977' 

1078- 

542 


TABLE    XXVIII. 

TRANSVERSE    STRAINS     IN    SPRUCE. 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-  > 

51 

52 

53 

54 

55 

56 

57 

58 

59 

60 

MENT.    ) 

DEPTH  1 

(in  inches),  f 

2-OI 

2-00 

1-99 

1-99  |  2-02 

1-99 

1-98 

2-01 

2-01 

2-02 

BREADTH  1 
(in  inches),  f 

I  -08 

i  i  -08 

I-  08 

1-08  i  i.  08 

i 

i  -06 

I  -08 

1-09 

1-07 

1-09 

PRESSURE 

(in  pounds). 

DEFLECTION  (in  inches). 

o 

•OOO 

OOO 

•OOO 

OOO 

•000 

•OOO 

•OOO 

•OOO 

•OOO 

•000 

5O 

•015 

•015 

•OIO 

•OIO 

•OIO 

•OIO 

•OIO 

•OIO 

•015 

•005 

100 

•025 

•025 

•O2O 

015 

•O2O 

020 

•O2O 

•020 

•  025 

•015 

150 

•035 

•035 

030 

•025 

•025 

•O25 

•025 

025 

•035 

•020 

200 

•040 

•045 

'035 

•030 

•035 

030 

•035 

•035 

C45 

030 

O 

•OOO 

•OOO 

•OOO 

OOO 

•OOO 

•000 

OOO 

•  000 

•000 

000 

200 

•040 

•045 

035 

-030 

•035 

•030 

•035 

•035 

•045 

•  030 

250* 

•050 

•055 

•045 

•040 

045 

•035 

040 

•040 

•055 

•040 

300 

•060 

•065 

•  050 

•  050 

055 

•  045 

•  050 

045 

•  070 

050 

350 

•070 

•075 

•  060 

•055 

•065 

•050 

•055 

055 

•  080 

•055 

400 

•075 

•085 

065 

•065 

-070 

•060 

•065 

•060 

•090 

-065 

O 

•OOO 

•OOO 

•OOO 

OOO 

•OOO 

•000 

•000 

•OOO 

•000 

•OOO 

400 

•075 

•085 

•065 

065 

070 

•  060 

•065 

060 

•090 

•  065 

450 

•080 

•095 

•070 

•070 

•080 

•065 

•070 

•  070 

•IOO 

-070 

500 

•090 

•  105  -080 

080 

•090 

•070 

•080 

•075 

•IIO 

•080 

55O 

•  IOO 

•115 

•090 

•  085 

•IOO 

•080 

•085 

•  080 

•120 

•090 

60O 

•  no 

•125 

•095 

•095 

•  IIO 

•085 

•095 

•090 

•  130 

•  IOO 

O 

•OOO 

•005 

•OOO 

•OOO 

-000 

•OOO 

•000 

•OOO 

•005 

•OOO 

60O 

•no 

•125 

•095 

•095 

•  IIO 

•  085 

•C95 

•090 

•135 

•IOO 

65O 

•  I2O 

•135 

•  105 

•  IOO 

•"5 

090 

•  IOO 

•095 

•  140 

•IIO 

TOO 

•  130 

•145 

•no 

•  I  IO 

•125 

•IOO 

•  105 

•105 

•  150 

-115 

750 

•135 

•155 

•  I2O 

•115 

•135 

•  105 

•115 

•IIO 

•  160 

•125 

8OO 

145 

•165 

•125 

•125 

•145 

•115 

•125 

•120 

175 

•135 

0 

•005 

•OIO 

•005 

•000 

•005 

•000 

•005 

•005 

•OIO 

•005 

80O 

•145 

.165 

•125 

•  125 

•145 

•115 

125 

•120 

•175 

•135 

85O 

•155 

•175 

•135 

•  130 

•155 

•I2O 

•  130 

•125 

•185 

•  140 

90O 

•  165 

•185 

•140 

•  140 

•165 

•  130 

•  140 

•130 

•195 

•150 

950 

•175 

•195 

•  150 

•145 

•175 

•135 

•145 

•  140 

•2IO 

•160 

1OOO 

•  185 

•210 

•  160 

•155 

.[85 

•145 

•155 

•145 

•225 

•170 

543 


TABLE    XX VI 1 1.— (Continued) 


TRANSVERSE     STRAINS     IN     SPRUCE 

LENGTH  i  •  6  FEET  BETWEEN  BEARINGS. 

See  Arts.  7O4  and  7O5. 


NUMBER  OF  i 

1 

EXPERI-  V 

51 

52 

53 

54 

55 

56 

57 

58 

59 

60 

MENT.    ) 

DEPTH  I 
(in  inches),  f 

2-01 

2-00 

1-99 

1-99 

2  -O2 

1-99 

1-98 

2-01 

2-OI 

2-02  , 

BREADTH  1 
(in  inches).  ) 

I  -08 

I  -08 

i  -08 

i  -08  i  i  -08 

i  -06 

I  -08 

,.09 

1-07 

1-09 

PRESSURE 
(in  founds). 

DEFLECTION  (in  inches). 

O 

•OIO 

•015 

•005 

•005 

•OIO 

•000 

•OIO 

•OIO 

•015 

OIO 

1000 

IQO 

•210 

•160 

•155 

•190 

•MS 

•155 

•145 

•22S 

•i/o 

1O5O 

•2OO 

•225 

•165 

160 

2OO 

150 

•160 

•150 

•235 

•180  | 

110O 

210 

•240 

175 

170 

210 

•160 

•170 

1  60 

•250 

•195  1 

115O 

230 

•255 

185 

•  1  80 

22O 

•165 

•175 

•170 

•270 

•205 

1200 

•245 

•27O 

195 

190 

•235 

"175 

•185 

•175 

•285 

215 

O 

•030 

•025 

•OIO 

•OIO 

•025 

•OIO 

OIO 

•OIO 

•030 

•O2O 

1200 

•250 

•275 

•  195 

190 

•240 

1  80 

•185 

175 

-285 

•22O 

125O 

270 

•290 

•205 

•2OO 

•255 

190 

•195 

•185 

•3OO 

•230 

13OO 

•295 

•305 

•215 

•2IO 

•270 

200 

205 

•195 

•320 

•245 

1350 

•325 

•325 

-230 

•220 

•28S 

•210 

•215 

•205 

•345 

•260 

140O 

•360 

•355 

•240 

•235 

•305 

•22O 

•225 

•215 

•365 

•275 

0 

070 

060 

•O2O 

•O2O 

040 

020 

•O2O 

•O2O 

•055 

•030 

1400 

•375 

•370 

245 

235 

310 

•22O 

•230 

•215 

•370 

•280 

145O 

•415 

400 

•255 

•250 

•330 

•235 

•240 

•230 

•390 

•295 

15OO 

430 

270 

265 

•355 

•250 

•255 

•245 

•415 

•320 

1550 

.  ,  .  . 

290 

280 

•265 

•275 

•260 

•350 

160O 

'  *?IS 

•  3O^ 

•28s 

•2Q^ 

•280 

•380 

0 

"OSS 

•040 

.... 

•040 

040 

•O4O 

O7O 

16OO 

•330 

•310 

•  290 

•300 

•290 

165O 

•375 

•335 

.... 

•  310 

•325 

•3IO 

1  TOO 

•  ^7^ 

•  335 

175O 

•  370 

BREAKING 

WEIGHT  (in 
pounds). 

1472- 

I536- 

1675- 

I7I7' 

1519- 

1653- 

1686- 

1800- 

1545' 

I6OO- 

544 


TABLE    XXIX. 


TRANSVERSE    STRAINS    IN   WHITE    PINE. 

LENGTH  i  •  6  FEET  BETWEEN  BEARINGS. 

See  Arts.    7O4  and  7O5. 


NUMBER  OF  ) 

EXPERT-  > 

61 

62 

63 

64 

65 

66 

67 

68 

69 

MENT.    ) 

, 

DEPTH  ) 

(in  inches).  ( 

1-02 

•99 

•99 

1-03 

1-02 

•99 

•99 

I-OO 

•99 

BREADTH  | 
(in  inches),  j 

I  -OO 

1-02 

1-02 

I-OO 

I-OI 

I  -01 

I-OI 

•99 

i  -02 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

O 

•OOO 

•OOO 

•OOO 

•000 

•000 

•000 

•000 

•000 

•000 

25 

•035 

•030 

•040 

•030 

•035 

•040 

•030 

•035 

•030 

50 

•070 

•  060 

•075 

•060 

•070 

•080 

•065 

•065 

•065 

15 

•IOO 

•095 

'US 

•090 

•IOO 

•115 

•095 

•095 

•095 

100 

•  130 

•125 

150 

•120 

•  130 

•  150 

•130 

•125 

•  125 

O 

•OOO 

•OOO 

•000 

•OOO 

OOO 

•OOO 

OOO 

•OOO 

•OOO 

10O 

•  130 

•125 

•150 

•120 

•  130 

•  150 

130 

•125  -125 

125 

•160 

•155 

•  185 

•150 

•165 

•185 

•160 

•  160 

•155 

15O 

•  190 

•185 

•220 

•180 

•195 

•22O 

•190 

190 

•185 

175 

•220 

•215 

•260 

•210 

230 

•250 

220 

•225 

•215 

2OO 

•250 

•245 

•295 

•240 

•  260 

•285 

250 

•255 

•245 

O 

•005 

•OOO 

•OO5 

-000 

•OOO 

•OIO 

•OOO 

•005 

•OOO 

200 

•230 

•250 

•295 

•240 

260 

•  290 

•  250 

•255 

•245 

225 

•280 

•280 

•335 

•270 

295 

325 

•285 

•285 

•280 

250 

•315 

•  310 

•380 

•300 

.330 

•365 

•320 

•320 

•310 

275 

•345 

•345 

425 

•335 

•365 

•410 

•350 

•355 

•345 

30O 

•380 

•375 

•470 

•365 

•405 

460 

•385 

•385 

•375 

O 

•OIO 

•005 

.... 

•005 

•OIO 

•030 

•005 

•OIO 

•OIO 

300 

•385 

•380 

.... 

•370 

.410 

•465 

•385 

•  390 

•380 

325 

•430 

•415 

.... 

•405 

•  460 

•520 

•430 

•425 

•415 

350 

•485 

•  450 

440 

•  e^£5 

.465 

•  4cc 

375 

•500 

•?*?*•* 

540 

DJO 

*rv-'D 

•  ^3^ 

TO  J 

•4QC 

Jjj 

T^V  J 

BREAKING 

WEIGHT  (in 
pounds). 

376-   385- 

300- 

382    356-  ;  350- 

349'   383- 

399- 

i 

545 


TABLE  XXX. 


TRANSVERSE    STRAINS    IN   WHITE    PINE. 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-   v 

70 

71 

72 

73 

74 

75 

76 

77 

78 

MENT.     J 

DEPTH  | 
(in  inches).  } 

i-53 

1-50 

1-52 

i-53 

I-5I 

1-49 

i-53 

1-52 

1-49 

BREADTH  I 
(in  inches).  ) 

1-02 

1-03 

1-03   1-02 

1-03 

I  -OI 

I  -02 

i  -02 

I  -OI 

PRESSURE 
(in  poundsy. 

DEFLECTION  (in  inches). 

O 

•ooo 

ooo 

•OOO 

•ooo 

ooo 

•ooo 

•000 

•000 

•000 

25 

•015 

OIO 

•OIO 

•015 

•OIO 

•015 

OIO 

•OIO 

•OIO 

50 

•025 

•O2O 

020 

-  -025 

•020 

•025 

•020 

•015 

•O2O 

75 

•035 

•030 

•030 

•035 

030 

035 

•030 

•  025 

•030 

100 

•045 

•035 

040 

•045 

•040 

•045 

040 

•035 

•O4O 

O 

•ooo 

000 

•000 

•  ooo 

•OOO 

•000 

•000 

•000 

000 

100 

•045 

035 

040 

•045 

•040 

•045 

•040 

•035 

•  040  i 

125 

•055 

045 

•050 

•055 

•050 

•055 

•045 

•040 

•050  ij 

15O 

•065 

•  050 

•  060 

•065 

•055 

•065 

•055 

•050  1  -060 

175 

•075 

•060 

•070 

075 

•  065 

•075 

•  065 

•  060   •  070 

2OO 

•085 

070 

•080 

•090 

•075 

•085 

•075 

•  070 

•080 

O 

ooo 

ooo 

•ooo 

•000 

000 

•ooo 

•ooo 

•000 

•ooo 

20O 

•085 

070 

-080 

•090 

•075 

•085 

•075 

•070 

•  080 

225 

•095 

•080 

•090 

•  IOO 

•085 

095 

•  080 

075 

•090 

25O 

•105 

•090 

•  IOO 

•no 

•095 

•105 

•090 

•085 

•  IOO 

275 

•115 

•100 

•no 

•  I2O 

•105 

•115 

•  IOO 

•095 

no 

30O 

•125 

105 

I2O 

•130 

no 

•125 

•105 

•105 

-125 

O 

•005 

•  oco 

•000 

•  ooo 

•ooo 

•ooo 

•ooo 

•ooo 

•ooo 

3OO 

•125 

•105 

•  I2O 

•  130 

•no 

•125 

•  105 

•  105 

•125 

325 

•135 

•115 

•130 

•  140 

•120 

135 

•115 

•115 

•135 

35O 

•145 

•125 

•140 

•  150 

125 

145 

•125 

•  I2O 

•145 

375 

•155 

135 

•145 

160 

135 

•155 

•135 

130 

•  160 

4OO 

•165 

•145 

•155 

170 

•145 

•165 

•145 

140 

•I/O 

0 

•005 

•005 

•005 

•005 

•ooo 

•000 

•000 

•000 

•ooo 

40O 

•165 

•145 

•155 

•170 

145 

•165 

•145 

•  140 

•  170 

425 

•175 

•150 

I65 

•  180 

150 

•175 

•  150 

•  145 

•180 

45O 

•185 

160 

•175 

•190 

160 

•185 

•  lt)O 

•155 

•190 

475 

•195 

•  170 

•I85 

•200 

•170 

•195 

•  170 

•165 

•200 

50O 

205 

180 

•195 

•2IO 

•180 

•205 

•175 

•175 

•210 

546 


TABLE   XXX.— (Continued^) 


TRANSVERSE     STRAINS     IN    WHITE    PINE. 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


1  NUMBER  OF  ) 

EXPERI-    > 

70 

71 

72 

73 

74 

75 

76 

77 

78 

MENT.    ) 

(in  inches),  f 

i-53 

1-50 

1-52 

i  53 

i-5i 

1-49 

i-53 

1-52 

i-49 

BREADTH  | 
(in  inches),  f 

I  02 

I  03 

1-03 

i  02 

1-03 

I  OI 

1-02 

1-02 

I  01 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

O 

005 

•005 

•005 

•005 

•000 

•005 

•000 

•005 

•ooo 

5OO 

•205 

•180 

•195 

•215 

•180 

•205 

•175 

•175 

•21O 

525 

•215 

•190 

•205 

•225 

•190 

•220 

•I85 

•I85 

•225 

550 

•225 

•  200 

•215 

•235 

•200 

•230 

•195 

•195 

•235 

575 

•235 

•2IO 

•225 

•250 

•210 

•240 

•205 

•205 

•250 

60O 

•245 

•220 

•235 

•2^0 

•22O 

•25O 

•215 

•215 

•260 

O 

•005 

•005 

•OIO 

•OIO 

•005 

•OIO 

•OO5 

•005 

•005 

GOO 

•245 

•22O 

•235 

•260 

•225 

•255 

•215 

•215 

•260 

625 

-260 

230 

•245 

-275 

•235 

•265 

-225    -225 

•27O 

650 

•275 

•240 

•255 

•290 

•245 

•280 

•235 

•235 

•285 

675 

•290 

•2<0 

•26$ 

•305 

•255 

•295 

•245 

•245 

•300 

700 

•305 

•260 

•275 

•325 

•265 

•310 

•255 

•255 

•32O 

O 

•O2O 

•010 

•OIO 

•025 

•OIO 

•020 

•OO5 

•005 

•015 

700 

•3*5 

•265 

•275 

•335 

•265 

•315 

•260 

•255 

•320 

725 

•345 

•275 

•290 

•275 

•335 

•275 

•265 

•340 

75O 

•445 

•290 

.300 

.... 

•285 

•355 

•285 

•280 

•365 

775 

•  310 

•315 

.... 

•305 

•375 

•3OO 

•295 

SOO 

•330 

•335 

•325 

•395 

•320 

•315 

O 

•  030 

•020 

•020 

•040 

•025 

•025 

800 

.... 

•  340 

•340 

•335 

•405 

•325 

•32O 

825 

•365 

.... 

•355 

•430 

•355 

.340 

85O 

SQO 

•380 

•  jet 

•7QO 

•  ^^ 

875 

•465 

•  37^ 

90O 

.  400 

O 

•O45 

9OO 

•4IO 

I 

925 

•  450 

H 

i 

BREAKING  ) 

I 

WEIGHT  (in  V 
Pounds).   ) 

753-  |  824 

877-  :  720 

874- 

854- 

869- 

947' 

773-  i 

547 


TABLE    XXXI. 

TRANSVERSE     STRAINS    IN     WHITE    PINE. 

LENGTH  i  6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-  V 

79 

SO 

81 

82 

83 

84 

85 

86 

87 

MENT.    J 

DEPTH  \ 
(in  inches).  ) 

2-11 

2-IO 

2-05 

2-09 

2-09 

2-08 

2-06 

2-09 

2-II 

BREADTH  \ 
(in  inches).  J 

I  04 

1-04 

i  03 

1-03 

1-03 

1-03 

1-04 

1-03 

1-03 

PRESSURE 
(in  pounds). 

DEFLECTION  (In  inches). 

1 

0 

•000 

•000 

•000 

•000 

•ooo 

•ooo 

•ooo 

•000 

•ooo 

5O 

•010 

•OIO 

•0!5 

•015 

•OIO 

015 

•OIO 

•OIO 

•  015 

too 

•O2O 

•020 

025 

•025 

•O2O 

•030 

•O2O 

•020 

•025 

15O 

•030 

•030 

040 

•035 

•030 

•040 

•  030 

•030 

•035 

200 

•035 

•040 

•050 

•040 

•035 

•050 

•040 

•040 

•040 

O 

•005 

•000 

•005 

•005 

•005 

•005 

•ooo 

•ooo 

•005 

200 

•035 

•040 

•050 

•040 

•035 

•050 

•040 

•040 

•040 

250 

045 

•045 

•O6o 

•050 

•045 

•060 

•045 

•050 

•050 

30O 
350 

•055 
•O6O 

•055 

•060 

•O7O 
•O8o 

•055 
•065 

•055 

•060 

•070 
•080 

•055 
•065 

•060 

•065 

•055 
•065 

400 

•070 

•070 

Ogo 

•075 

•070 

•090 

•070 

•075 

•070 

O 
400 

010 

•005 

OIO 

•OIO 

•OIO 

OIO 

•005 

•005 

•OIO 

450 
5OO 
55O 

•070 

•075 
-085 

•070 
•080 

-085 

•090 

IOO 

no 

•075 

080 
•090 

•070 

•075 
•085 

•090 

•  IOO 
•IIO 

•070 

•  080 

•090 

•075 

•080 

•090 

•070 
•080 

•085 

600 

•095 

•095 

•I2O 

•095 

090 

•120 

•IOO 

•  IOO 

095 

•105 

•IOO 

130 

•  105 

•IOO 

130 

•  IIO 

•  IIO 

•  I  'JO 

O 

•015 

OIO 

•015 

015 

•015 

•020 

•OIO 

•OIO 

•015 

60O 

•105 

•  IOO 

-130 

•105 

•TOO 

•130 

•IIO 

•IIO 

•  IOO 

650 

•  no 

•no 

140 

115 

•no 

I4O   -I2O 

•120 

•  IIO 

700 

•120 

•US 

•155 

•125 

•120 

•ISO    -130 

•  130 

•115 

75O 

•125 

•125 

•165 

•135 

•130 

•  160   •  140 

•140 

'1*5 

8OO 

•135    -130 

•i/5 

•145 

•135 

•170   -150 

•150 

•135 

548 


TABLE    XXXL— (Continued) 

TRANSVERSE     STRAINS     IN     WHITE     PINE. 

LENGTH  1.6  FEET  BETWEEN  BEARINGS. 

See  Arts.    7O4  and  7O5. 


NUMBER  OF  ) 

i 



EXPERI-  V 

MENT.    ) 

79 

SO 

81 

82 

83 

84 

I  85 

86 

87 

DEPTH   \ 
(in  inches),  f 

2-II 

2'10 

2-05 

2-og 

2-09 

2-08 

2-06 

2-09 

2-It 

BREADTH  I 
(in  inches),  f 

I  04 

I  04 

1-03 

1-03 

1-03 

1-03 

1-04 

1-03 

1-03 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches}. 

o 

•020 

•015 

•020 

•O2O 

•020 

•030 

015 

015 

O20 

8OO 

•135 

•130 

•175 

•145 

•135 

•170 

150 

•150 

135 

85O 

140 

140 

190 

'ISO 

'MS 

•180 

•160 

•  160 

•140 

90O 

150 

•145 

•200 

•160 

•150 

•195 

•170 

•170. 

•150 

950 

•155 

•155    -215 

•  170 

•  160 

-2  5 

•180 

180 

•155 

1OOO 

•I65 

•160   -230 

•  185 

•170 

•  220 

•190 

.  -I9O 

•I65 

0 

•025 

•020   -035 

•025 

•025 

•035    -O2O 

•025 

•O2i; 

1OOO 

•I6.S 

•160   -235   -185 

•170 

•220 

•190 

•195 

•I65 

105O 

•175 

•170   -250 

•195 

•180 

•235 

•200 

•205 

•175 

11OO 

•I  80 

•180 

•275 

•205 

•190 

•250 

•215 

•215 

•I85 

115O 

•I85 

•  190 

•22O 

•200 

•27O 

•235 

•230 

•195 

12OO 

•195 

•2OJ 

•240 

•2IO 

•295 

•255 

•245 

•205 

0 

•030 

•O25 

•045 

•030 

-060 

035 

•040 

•035 

1200 

•205 

•2OO 

•245 

•2IO 

•310 

•265 

•255 

•2IO 

125O 

210 

•  210 

.... 

•270 

•22O 

•340 

•285 

•275 

•22O  i 

1300 

•22O 

•22O 

... 

305 

•23O 

•310 

•335 

230 

135O 

•230 

•230 

.... 

•390 

•245 

.... 

.... 

•245 

1/1  on 

•240 

•240 

•260 

•26o 

0 

•040 

•  o^o 

'OJC 

•O^'s 

140O 

•245 

•245 

.... 



•270 



•265  | 

145O 

•26O 

•  260 

•  ^oo 

•  2QO  i 

150O 

•285 

•  280 

.... 

.... 

.... 

•315 

155O 

•  ^1^ 

•  160 

16OO 

•355 

BREAKING  ) 

WEIGHT  (in  V 
pounds).  ) 

1629- 

1536. 

1150- 

I383' 

1500- 

I280' 

1349- 

1303- 

1553- 

549 


TABLE    XXXII. 


TRANSVERSE     STRAINS     IN    WHITE    TINE. 

LENGTH  i  FOOT  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF  ] 

—  _  .  

EXPERI- 

278 

279 

28O 

281 

282 

283 

284 

285J  286 

MENT.       ) 

DEPTH     1 

(in  inches),  j 

•251 

253 

•258 

•498 

•502 

•S°3 

-748 

•746 

•747 

BREADTH  | 
(in  inches).  J 

I  -OOO 

I-OOO 

I-OOO 

i  -ooo 

I  -OOO 

I-OOO 

r  -ooo 

I-OOO 

I'OOO 

1 

PRESSURE 
(in  pounds). 

DEFLECTION 
(in  inches). 

PRESSURE 
(/* 

pounds). 

DKFLECTION 
(in  inches). 

PRESSURE 
(in 

pounds). 

DEFLECTION 
(in  inches). 

0 

•ooo 

ooo 

•000 

0 

•ODO 

•000 

000 

0 

•ooo 

ooo 

ooo 

1 

•019 

022 

016 

4 

•cog 

•on 

009 

10 

•007 

•009 

•007 

1 

•038 
•057 

044 
066 

•032 
•048 

8 
12 

•0,8 
•026 

•O2I 

•  032 

•018 
•027 

20 
30 

•014 

•O2I 

•018 
•027 

•014 

022 

4 

•076 

o38 

064 

16 

•034 

•042 

°35 

40 

•029 

•036 

•029 

5 

•095 

no 

•080 

20 

•043 

•052 

•044 

50 

•036 

•045 

•037 

6 

•114 

•132 

•096 

24 

•o5r 

062 

053 

6O 

•°43 

•054 

•045 

7 

•'33 

•'54 

.  •  112 

28 

•059 

O72          -062 

7O 

•050 

.063 

•052 

8 

•152 

•176       -128 

32 

•068 

•082 

•071 

80 

•057 

•072 

•059 

9 

•17'    ,    "9'3   ,      144 

36 

•076 

•092 

•079 

90 

.064 

•08  1 

•066 

1O 

•190       -220 

•160 

4O 

085 

•103 

088 

100 

•072 

•090 

•074 

11 

•209       -242 

•  176 

44 

•094 

•114 

•096 

110 

•079 

.099 

•082 

12 

•228       -264   •    -192 

48 

•102 

•125 

•105 

120 

•086 

•108 

•089 

13 

•247       -286       -208 

52 

•  1IO 

•136 

•114 

13O 

•  093 

•118 

•097 

14 
15 

•267   I    -308 
•286       -330 

•225 
•241 

60 

•iiS 
•127 

•I47 
•I57 

•123 
132 

140 
15O 

•  100 

•107 

•127 

•136 

I04 
•  112 

16 

•305   |    -352 

•257 

64 

•136 

•l67 

•141 

160 

•114 

-146 

•119 

17 

•324 

'374 

•273 

68 

•J45 

I   ?8 

•  150 

170 

-  121 

•127 

18 

•343 

-396 

•290 

72 

•'54 

•,89 

159 

18O 

•128 

•165 

*35 

19 
20 

•362 
•381 

•419 

•442 

•306 
-322 

76 

80 

-.63 
•172 

•'99 

•210 

•168 
•176 

19O 
20O 

•135 
I42 

•176 

•143 
152 

21 

•400 

•466 

••339 

84 

•181 

•221 

-185 

210 

•'49 

•200 

•161 

22 
23 

•419 
438 

•490 
•515 

356 

•373 

88 
92 

-191 

•200 

•232 
244 

'94 
203 

220 
230 

•156 

164 

•2I3 

•227 

171 
•181 

24 
25 

'457 

•477 

:567 

•390 
•407 

96 
100 

•209 
-219 

256 

268 

213 
225 

240 
250 

•171 
179 

•247 

•'95 

210 

26 

•497 

•594 

•425 

104 

•229 

•280 

237 

260 

-186 

233 

27 

28 

•518 

443 
•462 

108 
112 

•239 
•249 

•294 
•308 

•250 
•262 

270 
280 

•194 

•201 

29 
30 

583 

•482 
505 

116 
120 

•258 
•268 

•323 
•338 

•277 
296 

29O 
3OO 

•20-) 
•2l3 

31 
32 
33 

•606 
•629 
•654 

.... 

•11 

124 
128 
132 

•278 
288 
299 

•354 
•373 

317 

310 
320 

•228 
•239 
.251 

136 

310 

340 

•26S 

140 

321 

144 

•332 

148 

344 

152 

'357 

156 

371 

160 

•3*7 

550 


TABLE    XXXIII. 


TRANSVERSE     STRAINS     IN     HEMLOCK. 

LENGTH  i  6  FEET  BETWEEN  BEARINGS. 

See  Arts.  7O4  and  7O5. 


NUMBER  OF  J 

EXPERI-   V    88 

89 

90 

91 

92 

93 

94 

95 

96 

MENT.     \ 

DEPTH   I 
(/#  inches),  f 

1-07 

r  -08 

I  07 

1-07 

I  09 

I  -10 

1-09 

I  08 

I  •  II 

BREADTH  1 
(in  inches).  \ 

1-07 

I  06 

I  09 

i  -06 

I  05 

i  -06 

1-07 

1-08 

1-09 

PRESSURE 

DEFLECTION  (in  inches). 

(in  pounds'). 

O 

•  ooo 

•OOO 

•ooo 

000 

•000 

•000 

•000 

•ooo 

•000 

25 

•050 

050 

•030 

025 

•035 

025 

•040 

•035 

•040 

50 

•ogo 

085 

060 

050 

070 

050 

•080 

065 

075 

75 

•i  20 

•125 

•090 

•075 

100 

•  070 

•  12O 

•095 

•110 

100 

150 

165 

•115 

•105 

135 

•095 

•  160 

•125 

•145 

O 

•  OIO 

005 

005 

ooo 

ooo 

•005 

•ooo 

coo 

ooo 

100 

160 

•165 

•  1  20 

105 

•  140 

•095 

•160 

•125 

145 

125 

•190 

•210 

150 

130 

-170 

•  1  20 

200 

•155 

185 

150 

•225 

25O 

.185 

160 

•2OO 

•  140 

•240 

.185 

22O 

175 

265 

•295 

215 

•  190 

230 

•165 

•285 

•220 

•260 

200 

300 

340 

•245 

•220 

260 

190 

•330 

•250 

•  500 

0 

OIO 

015 

•005 

•OOO 

•005 

•005 

•OIO 

•O05 

•OIO 

200 

•305 

345 

•250 

•220 

265 

•185 

•330 

•250 

300 

225 

•340 

•400 

•285 

250 

•305 

•210 

.380 

-285 

•340 

250 

•380 

•455 

•315 

•2SO 

•235 

•430 

•385 

275 

-420 

•510 

350 

•310 

•260 

•480 

.... 

•430 

300 



•395 

345 

— 

•285 

•540 

475 

O 

025 

005 

.OIO 

•045 

•040 

30O 

.  . 

.... 

400 

350 

•285 

•570 

•490 

325 

.... 

390 

.... 

'315 

.... 

•545 

35O 

•350 

375 

••*8* 

40O 





J  J 

•445 

O 

•045 

400 

•46=; 

425 

*TWD 

•  530 

BREAKING 
WEIGHT  (in 

292- 

277   324 

350 

234   433 

3i3- 

,33 

348- 

pounds). 

TABLE   XXXIV. 


TRANSVERSE     STRAINS     IN     HEMLOCK. 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.   7O4  and  7O5. 


NUMBER  OF) 

EXPERI-  > 

97 

98 

100 

101 

102 

103 

104 

105 

106 

MENT.    ) 

DEPTH  ) 
(in  inches),  f 

1-56 

I  -60 

1-60 

i-59 

1.56 

I  -60 

i-54 

i'54 

1.58 

1-58 

BREADTH  1 
(in  inches).  ) 

1-04 

I  -06 

1-07 

1-03 

I-OI 

I  -08 

1-09 

i-  IT 

I  -08 

1-09 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

0 

•ooo 

•ooo 

•ooo 

•000 

•ooo 

•ooo 

•ooo 

•  ooo 

•000 

•ooo 

25 

•OIO 

•OIO 

•OIO 

•015 

•OIO 

•015 

•015 

015 

•OIO 

•015 

50 

•O2O 

•025 

•025 

•030 

•015 

•  030 

•030 

•  030 

•  025 

•  030 

75 

•030 

•035 

•035 

•040 

•  025 

•045 

045 

•045 

•040 

•  045 

100 

•040 

•050 

•050 

•055 

•035 

•060 

•060 

•055 

•055 

•055 

0 

•OOO 

•ooo 

•000 

•000 

•000 

•000 

•ooo 

•ooo 

•  ooo 

•ooo 

100 

•040 

•050 

•  050 

•055 

•035 

•060 

•  060 

•055 

•055 

055 

125 

•045 

•  060 

•  060 

•065 

•045 

•075 

•075 

070 

•070 

•070 

150 

•055 

•075 

•070 

•080 

•055 

•090 

•090 

•085 

•085 

•  080 

175 

•065 

•  085 

•  080 

•090 

•065 

•  105 

•  105 

•IOO 

•  IOO 

•090 

200 

•070 

•095 

•090 

•  105 

•075 

•115 

•125 

•115 

•115 

•105 

0 

•ooo 

•000 

•ooo 

•000 

•ooo 

•ooo 

•  ooo 

•ooo 

•000 

•ooo 

200 

•070 

•095 

•090 

-105 

•075 

•115 

125 

•115 

•115 

•  105 

225 

•080 

•105 

•  105 

•115 

•080 

•130 

•  140 

130 

•  130 

•120 

250 

•085 

•115 

•120 

•  130 

•090 

•145 

•155 

•145 

•140 

•  130 

275 

•095 

•130 

•130 

•145 

•095 

•160 

-170 

•  160 

•155 

•145 

300 

•100 

•  140 

•  140 

•160 

•105 

•175 

•  185 

•175 

•yo 

•155 

0 

•005 

•000 

•005 

•ooo 

•000 

•005 

•  005 

•coo 

•000 

•ooo 

3OO 

•IOO 

•  140 

•145 

•160 

•  105 

•175 

•190 

•175 

•  170 

•155 

325 

•no 

•155 

•155 

•175 

•115 

•  190 

•  205 

•  190 

•185 

•165 

350 

•I2O 

•  170 

•  170 

•  190 

•125 

•2IO 

•  225 

•  205 

•2CO 

•175 

375 

•  125 

•180 

•  I  80 

•205 

•135 

•225 

•240 

•22O 

•215 

•  190 

400 

•135 

•190 

•  190 

•22O 

•145 

•240 

•260 

•235 

•  230 

•200 

552 


TABLE    XXXIV.— (Continued) 


TRANSVERSE     STRAINS     IN     HEMLOCK 

LENGTH  i  •  6  FEET  BETWEEN  BEARINGS. 

See  Arts.    7O4  and  7O5. 


!  NUMBER  OF  ) 

EXPERI-   > 

97 

98 

99 

1OO 

1O1 

1O2 

1O3 

1O4 

105 

1O6 

MENT.    ) 

DEPTH   1 
(in  inches).  ) 

1-56 

I  -60 

I  -60 

1-59 

1-56 

I  -60 

i-54 

i-54 

1-58 

1-58 

BREADTH  1 
(in  inches),  ) 

1-04 

I  -06 

1-07 

1-03 

I-OI 

I  -08 

1-09 

i  -ii 

I  -08 

1-09 

PRESSURE 
(in  pounds). 

DEFLECTION  (in  inches). 

0 

•005 

•000 

•OIO 

•005 

•005 

•015 

•015 

•OIO 

•OIO 

•coo 

40O 

•135 

•190 

•  190 

•220 

•145 

•240 

•265 

•235 

•230 

•200 

425 

•145 

•205 

•205 

•240 

•150 

•255 

•285 

•255 

•250 

•215 

45O 

•  150 

•220 

'•220 

•255 

•160 

•275 

•305 

•270 

•265 

•23O 

475 

•160 

•230 

•23O 

•270 

•170 

•290 

•325 

•285 

•285 

•240 

500 

•170 

•240 

•245 

•295 

•175 

•310 

•345 

•305 

•305 

•255 

O 

•005 

•OIO 

•015 

•020 

•OIO 

•025 

•030 

•020 

•020 

•OIO 

500 

170 

245 

•245 

•300 

•180 

•315 

•355 

•310 

•305 

•255 

525 

-180 

•  260 

•260 

•340 

•190 

•335 

•380 

•330 

•325 

•  270 

55O 

•  iSS 

•275 

•275 

•200 

•355 

•405 

•350 

•345 

•285 

575 

•195 

•  300 

•290 

.... 

•210 

•380 

.430 

•370 

•365 

•300 

600 

•205 

— 

•305 



•22O 

-400 

460 



•385 

•315 

0 

-OIO 

•O25 

•015 

•045 

•060 

.... 

•030 

•O2O 

600 

•205 

•310 

.... 

•225 

•405 

•470 

.... 

•390 

•320 

625 

•215 

.... 

•325 

•235 

•435 

•500 

.... 

•415 

•335 

.  22^ 

•  "3AO 

•  24.  ^ 

•465 

•435 

•  3^0 

o5U 
675 

.240 

•360 

.... 

•265 

.... 

.... 

•460 

•370 

70O 

•26O 

— 

•380 



•290 

.... 

.... 

.... 

.... 

•385 

•O2O 

•O^O 

•  035 

TOO 

•  260 

•  ^o^ 

•3QO 

•  280 

•4(X 

1  Ah 

T  "iO 

•  2Q^ 

•425 

.  360 

•450 

BRFAKING  ) 

WKIGHT  (in  r 

777- 

575- 

700- 

548- 

727- 

651- 

650- 

600- 

687-  800  • 

pounds).  ) 

1     1 

553 


TABLE    XXXV. 


T  R  A  N  S  VE  R  S  E    STRAINS    IN     HEMLOCK 

LENGTH  1-6  FEET  BETWEEN  BEARINGS. 

See  Arts.    7O4  and  7O5. 


NUMBER  OF  ) 

EXPERI-     V 

107 

108 

109 

no 

111 

112 

113 

114 

115 

MENT.         J 

DEPTH      | 

(in  inches),  j 

2  -O2 

2   02 

2-00 

2-03 

2-01 

2-01 

1-99 

2-00 

2-03 

BREADTH    | 
(in  inches),  j 

1-03 

1-05 

I-03 

1-04 

1-05 

1-04 

1-02 

1-03 

1-05 

PRESSURE 

(in   founds). 

DEFLECTION  (in  inches). 

O 

•ooo 

•000 

•oco 

•ooo 

•000 

•000 

•ooo 

•ooo 

•ooo 

5O 

•015 

•OIO 

•OIO 

•OIO 

•015 

•015 

•O2O 

•OIO 

•OIO 

100 

•025 

•020 

O2O 

•020 

••025 

•030 

•030 

•020 

•025 

15O 

•035 

•035 

•  030 

•035 

035 

•045 

•045 

•035 

•°35 

2OO 

•045 

•045 

•045 

•050 

•045 

•055 

•055 

•045 

•045 

O 

•ooo 

•000 

•000 

•ooo 

•ooo 

•ooo 

•000 

•000 

•ooo 

200 

•045 

•045 

•045 

•050 

•045 

•055 

•055 

•°45 

•045 

250 

'055 

•060 

•°55 

•065 

•055 

-.065 

•065 

•055 

•°55 

300 
35O 

•065 
•075 

•070 

•085 

•065 
•075 

•075 
•090 

•070 

•085 

•075 
•085 

•075 
•090 

•065 
•«75 

065 
•075 

40O 

•085 

•095 

•083 

•IOO 

•095 

•095 

•IOO 

•090 

•085 

0 
40O 

•005 
•085 

•ooo 

•095 

•CDS 
•085 

•005 

IOO 

•005 
•095 

•  005 

•°9S 

•ooo 

IOO 

•000 

•090 

•000 

•085 

45O 

•095 

•105 

095 

•  115 

•  105 

•105 

•  no 

•  IOO 

•  IOO 

500 

•105 

•  I2O 

•105 

•  130 

•120 

•"5 

.  I2O 

•no 

•no 

55O 

•"5 

•130 

•"5 

•MS 

•135 

•125 

•  130 

•  I2O 

•  1  20 

6OO 

•125 

•145 

•125 

•155 

'MS 

•135 

•140 

•  130 

•130 

O 

•005 

•005 

•005 

•005 

•005 

•005 

•  005 

•005 

•005 

60O 

•125 

•>45 

•125 

•'55 

•  140 

•130 

130 

650 

•135 

•160 

•*3S 

•170 

•160 

'MS 

•150 

140 

•140 

70O 

•145 

•170 

•150 

•185 

•170 

•165 

•*5S 

•150 

75O 

•160 

185 

•165 

•200  . 

•185 

•  165 

•175 

•165 

•165 

800 

•170 

•200 

•175 

•220 

•205 

•175 

•190 

•180 

•180 

O 

•OIO 

•005 

•005 

•OIO 

•005 

•OIO 

•OIO 

•005 

•OIO 

80O 
85O 

•170 

•185 

•200 
•2I5 

•III! 

•220 
•240 

•205 
•225 

•175 
•185 

•190 

•200 

•180 
•190 

•180 
•'95 

9OO 

•200 

•230 

•195 

•260 

•245 

•215 

•205 

•2,0 

95O 

•215 

•2S0 

•2IO 

•285 

•265 

•2,0 

•235 

•215 

•225 

1OOO 

•230 

•275 

•220 

•3°5 

•220 

•2OO       j        .... 

•250 

O 
1000 

•025 
•240 

•025 
•280 

•OIO 
•22S 

.... 

•025 
•310 

•015 
•220 

•025 
•270 

•020 

•255 

!().,() 

•255 

'3IS 

•240 

•235 

•280 

1100 

•280 

•370 

•260 

115O 

•320 

•290 

BREAKING   ) 
WEIGHT   (in  > 
founds).      ) 

"54- 

,1,1. 

i,  Si- 

991-               1049- 

1099- 

1036- 

98S- 

1075- 

554 


TABLE     XXXVI. 

TENSILE    STRAINS    IN     GEORGIA    PINE. 
See  Arts.  7O4  and  7O6. 


NUMBER  OF  ) 

i                | 

EXPKRI-     v 

116       117 

118 

119 

12O 

121 

122 

123 

124 

MENT.          ) 

DIAMETER  { 

(in  inches).  \ 

•355 

•355 

•350 

•355 

•355 

•345 

•345 

•355 

•365 

BREAKING  ) 

Less  than 

Less    than 

WEIGHT  (in  > 
founds).     ) 

2005- 

I6OO- 

I300- 

2152- 

1400- 

1924- 

1091- 

1306- 

1268- 

TENSILE   STRAINS   IN   LOCUST. 


NUMBER  OF  ) 

EXPERI-   V   125  !  126 

127 

128 

129 

13O 

131 

132 

133 

MENT.     ) 

DIAMETER  I 

* 

(in  inches'),  f 

•355 

•345 

•305 

•305 

.300 

•300 

•300 

•300 

•300 

BREAKING  1 

More  than 

WEIGHT  (in  V 
pounds).  ) 

"37- 

2265- 

24OO- 

I592- 

2074- 

1561- 

2131- 

1799- 

2395- 

TENSILE   STRAINS   IN   WHITE   OAK. 


NUMBER  OF  ) 

EXPERI-  > 

134 

135 

136 

137 

138 

139 

140 

141 

142 

MENT.    ) 

DIAMETER  I 

(in  inches).  | 

•355 

•3^5 

•345 

•355 

•305 

•300 

•305 

.300 

•300 

BREAKING  ) 

WEIGHT  (in  v 
founds).  ) 

1908- 

1303- 

1182- 

2375' 

1700- 

1442- 

1003- 

1319- 

2205- 

555 


TABLE      XXXVII. 


TENSILE     STRAINS     IN     SPRUCE. 
See  Arts.   7O4  and  7O6. 


NUMBER  OF  ) 
EXPERI-  \\  143   144 

145 

146 

147 

148 

149 

15O 

151 

MENT.    )  |         ! 

DIAMETER  | 

(in  inches).  \  . 

•305 

•305 

•300 

•305 

•305 

•305 

•355 

•365 

•360 

BREAKING  J 

WEIGHT  (in  V 
pounds).  ) 

1573- 

I402- 

1560- 

1368- 

I385- 

1533- 

1882- 

2078- 

1600  • 

TENSILE    STRAINS   IN    WHITE   PINE. 


NUMBER  OF  )    1 
EXPERI-     V      152 

153 

154 

155 

156 

157 

158 

159 

160 

MENT.          )     1 

• 

(in  inches),  f 

•395 

•365 

•365 

360 

-365 

•355 

•350 

•365 

•360 

BREAKING  ) 

WEIGHT  (in  V 
pounds).     ) 

1363- 

H57 

1127- 

1316' 

'431 

1487- 

1192- 

1024- 

1400 

TENSILE    STRAINS   IN   HEMLOCK. 


NUMBER  OF  ) 



,  — 

EXPERI-     >- 

161 

162 

163 

164 

165 

MENT.         ) 

DIAMETER  } 

(in  inches),  f 

•365 

•355 

•345 

•36o  j    -355 

BREAKING  ) 

WEIGHT  (in  V 
fiounds).     \ 

645- 

897- 

864- 

999.       863- 

166 

167 

168 

169 

•355 

•350 

•335 

•355 

726- 

895- 

809- 

977- 

556 


TABLE   XXXVIII. 

SLIDING   STRAINS   IN    GEORGIA   PINE. 
See  Arts.  7Q4-  and  7O6. 


^  '  — 

NUMBER  OF    ) 

\ 

EXPERI-          >• 

170 

171 

172    !    173    |    174 

175 

176 

177 

178 

MENT.             ) 

! 

DIAMETER     ) 
(in  inches),    f 

•525 

•520 

•530 

•530 

•  520 

•520 

•525 

•520 

•530 

LENGTH        I 
(in  inches),    j 

1-065 

Broke 
in  two. 

I-O2O 

I  -OIO 

Broke 
in  two. 

1-040 

1-020     I-OI5 

1-050 

BREAKING     ) 
WEIGHT  (in    > 

1546- 

,1295- 

I4II- 

1571- 

I28l- 

1347- 

I520- 

1401  • 

1247- 

pounds).       ) 

SLIDING   STRAINS   IN   LOCUST. 


NUMBER  OF  ) 
EXPERI-   V 

MENT.     ) 

179 

180 

181 

182 

183 

184 

185 

186 

187 

DIAMETER  I 
(in  inches).  \ 

•530 

•530 

•535 

•525 

•525 

•530 

•535 

•525 

•525 

LENGTH   I 
(in  inches),  f 

"735 

•730 

•715 

•745 

Broke 
in  two. 

•760 

•730 

•715 

•760 

BREAKING  ) 

WEIGHT  (in  V 

1490- 

1236- 

1533- 

1192- 

I492- 

I758-  !  I403' 

I33I- 

1483- 

pounds).   ) 

SLIDING   STRAINS   IN   WHITE   OAK. 


NUMBER  OK  1 

| 

EXPERI-   > 

188  i  189 

19O 

191 

192   193 

194 

195 

196 

MEXT.    ) 

DIAMETER  ) 
(in  inches),  f 

•530 

•525 

•535 

•540 

•535 

•530 

•530 

•535 

•530 

LENGTH   i 
(in  inches).  \ 

•730 

•755 

•740 

•750 

•750 

•740 

•725 

•725 

•730 

BREAKING  ) 

WEIGHT  (in  V 

1308- 

1801-  1834- 

1502- 

1701- 

1359' 

1667- 

1321- 

1399' 

pounds).   ) 

557 


TABLE   XXXIX. 


SLIDING     STRAINS     IN     SPRUCE. 
See  Arts.   7O4  and  7O6. 


NUMBER  OF    ) 
EXPERI-       >• 

MENT.            ) 

197 

19S 

199 

200 

201 

202 

2O3 

204 

205 

DIAMETER     j 
(in  inches),    j 

•565 

•535 

•550 

•525 

•550 

•550 

•545 

•550 

•550 

LENGTH       \ 
(in  inches).     ) 

I-OIO 

•  990 

I-OIO 

I-OIO 

1-030 

1-020 

1-005 

I-OIO 

-990 

BREAKING     j 
WEIGHT  (in    > 
founds).      ) 

988- 

770- 

1130- 

882- 

927- 

976- 

1043- 

838- 

902- 

SLIDING   STRAINS   IN    WHITE   PINE. 


NUMBER   OF     ) 

1 

i 

EXPERI-       v 

206 

207       208 

209   :   210 

211       212 

213 

214 

MENT.            ) 

I 

! 

DIAMETER      ) 
(in  inches),    f 

•540 

•545 

•555 

•545 

•545 

•545 

•550 

•545 

•545 

LENGTH       1 

(in  inches),     f 

•995 

I-OOO 

•990 

I-OIO 

i  -025 

1*005 

i-oio 

i  -040 

'995 

BREAKING     ) 

WEIGHT  (in    V 
pounds).       ) 

730- 

907- 

792- 

803- 

842- 

8co- 

881- 

852- 

885- 

SLIDING    STRAINS   IN   HEMLOCK. 


NUMBER  OF    )  I 
EXPERI-      H   215 

216 

217 

218 

219 

220 

221 

222 

223 

MENT.             ) 

DIAMETER     1 
(in  inches),    f 

•540 

•540 

•545 

'540 

•530 

•540 

•540 

•535 

•530 

LENGTH       ) 
(in  inches'),    f 

•995 

I-OIO 

•995 

Broke 
in  two. 

1-025 

1-015 

'995 

I  -010 

•99° 

BREAKING     ) 
WEIGHT  (in    V 

607- 

702- 

620- 

796- 

700- 

674- 

ss&. 

627- 

530- 

founds).      ] 

! 

558 


TABLE    XL. 

CRUSHING   STRAINS    IN    GEORGIA   PINE. 
See  Arts.   7O4  and  7O7. 


NUMBER  OF    ) 
EXPERI-       V 

MENT.             ) 

224 

225 

226 

227 

228 

229 

230 

231 

232 

DIAMETER     ) 
(in  inches),    f 

•515 

•515 

•520 

•52O 

•505 

•515 

•510 

•500 

•515 

LENGTH      ) 
(in  inches),    f 

1-035 

1-025 

I  -040 

1-035 

T-035 

•505 

•515 

•505 

•510 

BREAKING     ) 
WEIGHT  (in    V 
pounds}.      \ 

2010- 

1878- 

2061  • 

1735- 

2304- 

2002- 

1845- 

I705- 

2141- 

CRUSHING   STRAINS    IN    LOCUST. 


NUMBER  OF    ) 
EXPERI-       >• 

MENT.            ) 

233 

234 

235 

236 

237 

238 

239 

240 

241 

DIAMETER      j 
(in  inches),     j 

•520 

•520 

•520 

•525 

•530 

•520 

•525 

•515 

•520 

LENGTH       ) 
(in  inches),    f 

1-055 

1-020 

1-035 

1-045 

-490 

•515 

•500 

•490 

•495 

BREAKING      ) 

WEIGHT  (in   \ 
pounds},      ) 

2338- 

239I- 

2547- 

2539- 

2695' 

2500- 

2495' 

2374- 

2672- 

CRUSHING   STRAINS   IN   WHITE   OAK. 


NUMBER  OF   ) 
EXPERI-       V 

MENT.            ) 

242 

243 

244 

245 

246 

247 

248 

249 

25O 

DIAMETER      | 
(in  inches),    j 

•525 

•530 

•520 

•530 

•525 

•520 

•525 

•520 

•515 

LENGTH       | 
(in  inches}.    \ 

1-035 

1-035 

1-035 

1-030 

1-035 

•505 

•500 

•485 

•515 

BREAKING     ) 
WEIGHT  (in   > 
pounds),      j 

1546- 

1978- 

1992- 

1455- 

1989- 

1650- 

2116- 

I387- 

1376- 

559 


TABLE   XLI. 

CRUSHING     STRAINS    IN     SPRUCE. 
See  Arts.  7O4  and  7O7. 


NUMBER  OF    ) 
EXPERI-       V 

WENT.            ) 

251 

252 

253 

254 

255 

256 

257 

258 

259 

DIAMETER     \ 
(in  inches),    f 

•535 

•535 

•535 

•535 

•535 

•530 

•540 

•530 

•530 

LENGTH       1 
(in  inches),     f 

1-035 

1-025 

1-040 

i  -030 

•490 

•480 

•485 

•495 

•490 

BREAKING     ) 
WEIGHT  (in    V 
Jouneis).       } 

1692- 

I7I5- 

1611  • 

1633- 

1871- 

1818- 

1812- 

1855- 

1832- 

CRUSHING   STRAINS    IN   WHITE    PINE. 


NUMBER  OF    ) 

I 

i              1              1 

EXPERI-       V     26O 

261 

262 

263 

264 

265    i    266 

267 

268 

MENT.               ) 

1 

1 

DIAMETER      1 
(in  inches},     f 

•540 

•525 

•535 

•515 

•530 

•535 

•525 

-510 

•540 

(in  inches),     f 

1-035 

1-035 

i  -040 

1-040 

1-030 

•495 

•49° 

•50£ 

•495 

BREAKING     ) 

I 

WEIGHT  (in    > 
pounds).      ) 

1454' 

I536- 

1473- 

I322- 

1297- 

1503- 

1624- 

1353- 

1540- 

CRUSHING   STRAINS   IN   HEMLOCK. 


NUMBER  OF   )  1 
EXPERI-       U   269 

MENT.             )  1 

270 

271 

272 

273 

274 

275 

276 

277 

DIAMETER     [ 
(in  inches)     ) 

•520 

•520 

•525 

•530 

•530 

•520 

•520 

•525 

-520 

LENGTH      1 
(in  inches),    f 

1-035 

1-030 

1-030    1-030 

•480 

•525 

•520 

•495 

•490 

BREAKING     ) 
WEIGHT  (in    \ 
founds).      \ 

II37' 

1178- 

1130-     1150- 

1150- 

1334' 

1290- 

I3I7- 

1320- 

560 


TABLE    XLII. 

TRANSVERSE      STRAINS. 

BREAKING  WEIGHTS  (in  pounds)  PER  UNIT  OF  MATERIAL   =  B. 

See  Arts.    7O4  and  7O5. 


i 

w 

• 

w 

w 

M 

^ 

2 

• 

2 

2  ^  • 

2; 

^                  jj 

s/ 

PU:  ' 

w  ~ 

t&~r+ 

Q.  -„ 

M 

C/5     M 

**'     M 

U    M 

Cj   ^ 

I-JH   *™* 

0  M       0  M 

o 

3  x 

U    X 

W   x 

5  x 

S   X 

5  x 

w  x 

W   x 

W    X 

•J   X 

O-. 
P4  - 

•J" 

^i 

FW  - 

CL,  ^ 

r"  ^ 

H 

&; 

w  ~     i  w  ~ 

w^ 

W 

O 

^ 

J 

^ 

* 

ffl 

i-U  M 

s 

950-       664- 

686- 

576- 

604- 

540- 

578- 

504- 

563. 

381. 

491- 

439' 

1023-  |    555- 

533' 

654- 

650- 

569- 

616- 

569- 

536- 

358- 

339" 

415- 

758- 

1406- 

660- 

533- 

574' 

627- 

480- 

590- 

425' 

415- 

409- 

459' 

95i- 

1286- 

626- 

694- 

598- 

642- 

576- 

482- 

492- 

461- 

337- 

370- 

945" 

1283- 

476- 

632- 

538- 

552- 

542- 

595- 

533- 

300- 

473- 

396' 

1014- 

1156- 

567- 

637- 

652- 

630- 

566- 

609. 

460- 

540- 

377- 

418- 

993- 

1342- 

S36' 

702- 

660- 

637- 

564- 

582- 

489- 

394- 

402-  \  410- 

785- 

530- 

501- 

593- 

614- 

654- 

619- 

643- 

463- 

302- 

365- 

383- 

949' 

1212- 

664- 

627. 

592- 

572- 

639- 

552- 

542- 

4i5- 

408- 

398- 

I28l- 

529- 

722- 

642- 

576- 

470- 

952- 

AVERAGE  BREAKING  WEIGHTS,  =  B. 

930- 

1061- 

578- 

637- 

612- 

600- 

576- 

570. 

500- 

396- 

407- 

4IO' 

561 


TABLE    XLIII. 


DEFLECTION. 

VALUES    OF    CONSTANT,'^. 
See  Arts.  7O4  and  7O5. 


w 

£-• 

1  M 

I'M 

O  " 

H  X 

'RUCE. 

'x  i". 

&£ 

w 

2 

w  x 

H  - 

H 

K 

M 

w  x 

i? 

c? 

1—  \  M 

K-M 

w  M 

Cfl%H 

x  M 

£  M 

S  M 

ffi  M 

0 

* 

^ 

* 

5155- 

4983- 

2599" 

3649- 

3329- 

2577' 

3088- 

2746. 

2484. 

2083- 

3112- 

2316- 

6302- 

4645- 

2033- 

3640- 

2909. 

22c;9- 

3378- 

3191- 

2658- 

1859. 

1986- 

1958. 

5199- 

5616- 

2360- 

2215. 

2679  ' 

2962- 

2806- 

2891- 

2130- 

2667. 

2077- 

2386- 

5555- 

5000- 

2057- 

3867. 

2965  ' 

3105- 

3124- 

2624- 

2489. 

2868- 

2940- 

1822- 

5498- 

5442- 

1704- 

3384- 

2766- 

2539- 

2940- 

3176- 

2581-  2259- 

3052- 

1988- 

6007- 

4998- 

I837- 

2932- 

329I- 

3382- 

2850- 

2932- 

2040- 

3056- 

1606- 

2190- 

6007- 

5239- 

1907- 

4004. 

3302- 

3I27- 

3257- 

3101- 

2371- 

1847. 

1660- 

2184- 

4807- 

4312- 

1842- 

3572-  3344' 

3I91' 

3267- 

3225- 

2323. 

2441- 

1725- 

2294- 

6336- 

5239- 

2253- 

3678-:  3714- 

2176- 

3338- 

2919. 

2509- 

1895. 

1683. 

2152- 

5033- 

I930- 

3967- 

3387- 

2682- 

1864. 

4952- 

AVERAGE  VALUES  OF  CONSTANT,  F. 

5652- 

5042- 

2052- 

3491' 

3I6, 

2804- 

3Il6- 

2978- 

2398- 

2331- 

2170- 

2143- 

1 

562 


TABLE   XLIV. 

TENSILE       STRAINS. 

BREAKING  WEIGHTS  (in  pounds)  PER  SQUARE  INCH  OF  SECTIONAL  AREA,  == 
See  Arts.   7O4  and  7O6. 


GEORGIA 
PINE. 

LOCUST. 

WHITE 
OAK. 

SPRUCE. 

WHITE 
PINE. 

HEMLOCK. 

20257- 

11487- 

19277- 

21530- 

11123- 

6164- 

•/•••• 

24229 
21790- 

12453- 
12644- 
23995. 

19189- 
22069- 
18724- 

11057- 
10771- 
12929- 

9062- 
9242- 
9815- 

21742- 

14144- 
20582- 

29341- 
22084- 

23268  • 
20400- 

18957- 
20982- 

13676- 
15023- 

8719- 
7335- 

11671- 

30147- 

13728- 

19014- 

12389- 

9302- 

I3I95- 

25451- 

18660- 

19860- 

9786 

9178- 

12118- 

33882- 

31194- 

15719- 

T3754- 

9871- 

AVERAGE  WEIGHTS  PRODUCING  RUPTURE,  =  T. 

16244- 

24801  • 

19513- 

19560- 

12279- 

8743- 

563 


TABLE    XLV. 

.SLIDING     STRAINS. 

BREAKING  WEIGHTS  (in  pounds)  PER  SQUARE  INCH  OF  SLIDING  SURFACE,  =  G. 
See  Arts.  7O4  and  706. 


GEORGIA 
PINE. 

LOCUST. 

WHITE 
OAK. 

SPRUCE. 

WHITE 
PINE. 

HEMLOCK. 

880- 

1218- 

1076- 

55i- 

433' 

360- 



1017- 

1447- 

463- 

530- 

410- 

831- 

1275- 

1474. 

647- 

459' 

364- 

934" 

970- 

1181- 

530- 

464- 







1349- 

521- 

480- 

410- 

793" 

1389- 

1103- 

554' 

465- 

392- 

904- 

ii43- 

1381- 

606- 

505- 

329- 

845- 

1129- 

1084- 

480- 

479' 

369- 

7i3- 

1183- 

1151- 

527- 

520- 

322- 

AVERAGE  RESISTANCE  TO  RUPTURE  PER  SQUARE  INCH,  =  G. 

843- 

1165- 

1250- 

542- 

482- 

369- 

564 


TABLE   XLVI. 

CRUSHING      STRAINS. 

CRUSHING  WEIGHTS  (in  founds)  PER  SQUARE  INCH  OF  SECTIONAL  AREA,  =  C. 
See  Arts.   7O4  and  7O7. 


GEORGIA 

PlNE: 

LOCUST. 

WHITE 
OAK. 

SPRUCE. 

WHITE 
PINE. 

• 

HEMLOCK. 

9649- 

11009- 

7142- 

7527- 

6349- 

5354- 

9015- 
9705- 
8170- 

11259- 

H993- 
11729- 

8966- 
9380- 
6595- 

7629- 
7166- 
7264. 

7095- 
6552- 
6346- 

5547- 
5220 
5240- 

II503- 

12216 

9188-      8323- 

5879-      5213- 

9611- 

11772- 

7769- 

8240 

6686- 

6281- 

9032- 

11525- 

9775- 

7912- 

7502- 

6074- 

8683- 

11396-     6531- 

8408  • 

6623- 

6084- 

I0278- 

12582-     6606- 

I 

8304- 

6724- 

6216 

AVERAGE  RESISTANCE  TO  CRUSHING  PER  SQUARE  INCH,  =  C. 

9516- 

11720- 

7995- 

7864- 

6640- 

5692- 

565 


DIRECTORY, 


DIGEST  OF  THE   PRINCIPAL  RULES. 


BELOW  may  be  found  the  numbers  of  such  formulas,  arti- 
cles, figures  and  tables  as  are  particularly  applicable  in  any 
given  problem. 

By  reference  to  these,  the  rules  needed  in  any  certain 
case,  occurring  in  practice,  may  be  more  readily  found  than 
by  either  the  index  or  table  of  contents. 


LEVERS — WOOD. 

'  Strain  at  wall, 

"  any  point,      .          Figs.  27,  28,  33,  (4®.), 
Size  when  at  the  point  of  rupture, 

"     to  resist  rupture  safely, 
f  Weight,    .         .*       . 

a5  I  Length, 

|  -I  Breadth, 

E     Depth, 

Deflection,        .         .         .        .        . 


.(6.) 


(19. \  (36.} 
.    (123.) 

(127.) 
.     (128) 
(129) 
(121.) 


~o 
>% 

I 
I 

"5 
3 


DIRECTORY.  567 

f  Strain  at  wall,  .  .  .     (75.) 

"       "  any  point,          .         .  Fig.  46,  (76.) 

Size  when  at  the  point  of  rupture,      .         .     (18.) 
1       '•     to  resist  rupture  safely,  .         .        (20.),  (77.) 
"     at  any  point  to  resist  rupture  safely,       (77.) 

[  Shape  of  lever, Fig.  47 

Weight, .  (186.) 

Length, (187.) 

|  \    Breadth, (188.) 

£       Depth, (189.) 

_  Deflection,  .         .  (140) 


Strain  at  wall, Figs.  45,  51 

"  any  point,      .  .Figs.  45,  48,  50,  51 

Size       "     "         " Art.  227 

Shape  of  lever, Figs.  31,  49 

Depth  at  any  point, (80.) 


LEVERS— ROLLED-IRON. 
Load  at  end.     Flexure.      Weight,    .  .  .  . 

Size,     .         .         .         .  (224), 


Load  uniformly  distributed.     Flexure.      Weight,  ...    (230) 

Size,. 


;68 


DIRECTORY. 


1 


SINGLE  BEAMS — WOOD. 

'  Strain  at  middle,        .         .         . 

"       "     any  point,    .... 

Size  when  at  the  point  of  rupture,  (9.),  (11.) 
44     to  resist  rupture  safely, 
"     at  any  point  to  resist  rupture  safely, 
| -I  Weight, 

Length,      '.".*.". 

Breadth, 

Depth,       *.       Y      '. 

Constant    B,          .         .         .         . 

i  Pressure  on  each  support, 
['Weight,          ....         .         . 

g  j  Length,       .         .         .         . 

J  Breadth,         .      \ .        . 

Depth, 

Deflection, 


.       (9.) 
Fig.  29 

(12.),(14.) 

.        .(21.) 

.     (37.) 

.        .(13.) 


.(11.) 


..(10.) 

(<?•),  (4-) 

(122.) 


(125.) 
(126.) 
(120.) 


E 

J    |J 
b 

X 


"  Strain  at  middle,       .  ...     (44-) 

"       "  the  load, Art.  190 

-  any  point,  .  Figs.  34,  35.  (44-),  (45.) 

Size  when  at  the  point  of  rupture,      .         .         .  (16.) 

"     to  resist  rupture  safely,  (23.),  (46.),  (47.),  (48.), 

(49.),  (50.) 


El 

3      I 


f       f  Strain  at  middle, 

44       "     any  point,     . 
|  i  Size  when  at  the  point  of  rupture, 
to  resist  rupture  safely, 

I-EH      | 

Shape  of  beam,  .... 
(  Pressure  on  each  support,    . 

['Weight, 

g      Length,  .         .  .... 

I  -j  Breadth, 

Depth, 

Deflection,          .... 


Art.  59,  (72.) 

Fig.  42,  (71.) 
.        .      (17.) 


-    Figs.  43,  44 

.    (3.)t  (4) 
.  (131.) 

?.) 


.  (135.) 


DIRECTORY.  569 

f         r  Strain  at  any  point,     Figs.  36,  37,  (53.),  (54.),  (55.) 
"  locations  of  weights,  Art.  (53,  (51.),  (52.) 
Size  to  resist  rupture  safely,  (30.),  (31.),  (56.),  (57.) 
.      f  f    I    Strain  at  any  point,  Fig.  38,  (61.),  (62.), 

-i  J 

•2      1 


W 

•o 

rt        -< 

0         I 

—  o 


"  locations  of  weights,      (58.),  (59.),  (60.) 


I  Size  to  resist  rupture  safely,          (65.  \  (66.),  (67.) 


~  f  Rupture.      Strain   at  any  point,    Figs.  52,  53,  (<$-*•)»  (95-), 


1  1 

31  I  Rupture.      Size  to  resist  safely, 

|  I  Flexure.  "      "         <4  "...    (189.  \  (193.) 


Loaded  (  Rupture.      Strains,  Figs.  39,  4°,  41,  Arts.  196  to   203, 

210,  211 

SINGLE   BEAMS  — ROLLED-IRON. 

f  General  rule,  .......  (216.) 

t  |  Weight, (217.) 

|  -!  Length, (218.) 

E  Deflection,  .  ....  (219.) 

(  Moment  of  inertia>     , (220.) 


Load  at  any  point.     Flexure.     Weight,  .... 

"    '!    "     -  Size,  .         .         .        (221.),  (222.) 

Load  uniformly  distributed.     Flexure.     Weight,      '.  .  .        (228.) 

Size, 


5/0 


DIRECTORY. 


I 


FLOOR   TIMBERS  —  WOOD. 

General  rule,  ....... 

Dwellings,  assembly  rooms,  etc.,      .         .         . 

General  rule,  ....... 

f  General  rule,       ....      (142.), 

I   Distance  from  centres,  .  I.  to  IV.,  (144), 
\   Length,        ....... 

\   Breadth,  .....         .         . 

(  Depth,         .  .      ...... 

f  General  rule,  .....  (U8.), 

\   Distance  from  centres,  V.  to  VIII.,  (150.), 
Length,   ....... 

Breadth,      ....... 

I  Depth,     .  ..... 

Solid  floors  of  wood,         XXL,  (310.),  (311.), 


(25.) 

(141-) 

(143.) 
(306.) 
(145.) 
(146.) 


(307.) 

(151-) 
(152.) 
(153.) 

(312) 


f  Rupture.    General  rule, 
!   „•    f  General  rule, 
I  j  |  J   Dwellings,  etc., 
j  UH    j  First-class  stores, 


.        .  .       (27.) 

.        .        .        .        (156.) 

IX.  to  Xn.,  382,  (308.) 

XIII.  to  XVI.,  383,  (309.) 


With  one  header (29.) 

"      two  headers,  (32.),  (33.),  (34.),  (35.),  (92.).  (93.) 
"      three     "  ....    (97.),(106.) 

g-    [  General  rule,  ....     (157.),  (161.) 
||  j   Dwellings,  etc.,  .         .  (158.),  (162.) 

I  First-class  stores,    .         .         .      (159),  (163.) 
(  General  rule,    .    (164.).  (107.),  (170.),  (174), 

(179.),  (183.),  (186.) 
rgl   ,    Dwellings,  etc.,  (165.),  (168.),  (175.),  (180.), 

g|  (184\(187.) 

|  First-class  stores,  (166),(169.\  (176),  (181.), 

(185.),  (188.) 

S!j    [  General  rule,  .      Figs.  55,  S^(190),  (194) 

1 1    i  Dwellings,  etc.,  .         .  (191.),  (195.) 

«    [  First-class  stores,  .         .         .       (192.),  (196.) 
Girders.     Rupture.    General  rule,         .         .         .  Art.  (37 


X      * 

o 


DIRECTORY.  571 

FLOOR    REAMS — ROLLED-IRON. 

|   f  ,j    f  General  rule (234.) 

Dwellings,  etc.,  .         .         .        XVIII.,  (236.\ 


J   i  E    |   First-class  stores,    .  XIX., 

fa    I 

2    f  d    f  General  rule,        ..... 

|   Dwellings,  assembly  rooms,  etc.,  .         .         .      (248.) 


E    [ fe    I  First-class  stores,         .         .         .         .         .  (249.) 

f        f       ,.:    f  General  rule, (250.) 

gl  -\  Dwellings,  assembly  rooms,  etc.,  .  (251.) 

~    L  First-class  stores,        .         .         .         .  (252.) 

f  General  rule,  .         ....  (253.) 
•2%  j  Dwellings,  assembly  rooms,  etc., 


'    1  First-class  stores,        .     (255. \  (257.),  (259.) 
f  General  rule,  ....      Art.  531 

(A         I 

£.£  J  Dwellings,  assembly  rooms,  etc.,     .     (260. )\ 


First-class  stores,  . 


FRAMED   GIRDERS. 

Proportionate  depth, (294-) 

Number  of  bays,   ........     (295.) 

Strains  in  a  framed  girder,         ....         Pigs.  93,  94 

"  diagonals, (296.) 

Tensions  in  lower  chord,  ......         (297) 

Areas  of  cross-section  in  lower  chord,         .         .         .     (299) 
"       "  "       "  upper      "          .         .     (301),  (303) 

Unsymmetrical  load,  divided  between  the  two  supports, 

Figs.  96,  97,  (304.),  (305.) 


572 


DIRECTORY. 


TUBULAR   IRON   GIRDERS. 

Load  at  middle.  Area  of  flange,         .          (264),  (®65) 

"      "any  point.  "        "         "  (266.) 

>s%  f  General  rule,  "  "  "  .  .  .  .  (267 .) 
o  o  1  \  Banks  and  assembly  rooms.  Area  of  flange,  (274-) 
J'l|  !  First-class  stores,  "  "  t4  (275.) 

Thickness  of  web  to  resist  shearing,         .         .         .         (268.) 

Weight  of  girder, (270.),  (271.) 

Economical  depth  of  girder,     ....     (276.),  (277.) 

CAST-IRON   GIRDERS. 

Load  at  middle.     Breaking  weight, (278) 

••     »       -         Safe  area  of  flange,    .  (279.) 

"     "  any  point.     Breaking  weight,        ....     (281.) 
Safe  area  of  flange,         .         .         .          (282.) 
Two  concentrated  loads.     Safe  area  of  flange,      .         (285),  (286) 
Safe  area  of  flange  at  middle,          .         .         .         (280.) 
"      "       "        "       "  any  point,         .         .         .     (283.) 
"    depth  at  any  point,  .         .         .         .         (284.) 

Arch  girder,  safe  area  of  tie-rod,       .         .         .     (287.) 
safe  diameter  of  tie-rod,     .     (288.),  (289.) 
.  Brick  arch,      "  "  «        «<  t  (290.) 

ROOF   TRUSSES. 

Comparison  of  designs, Art.  658 

Strains  derived  graphically,   ..         .       Arts.  660  to  668,  679 
Horizontal  and  inclined  ties,         .    Fig.  125,  Arts.  669  to  671 

Designing  a  roof, Arts.  672,  676 

Load  upon  a  roof,          .         .         .         .     Arts.  673,  674,  675 
Load  upon  each  supported  point  in  a  truss,    .  Art.  677 

"      "  the  tie-beam,      Art.  678 

Measuring  the  strains,  as  in  force  diagram,    .  Art.  680 

Arithmetical  computation  of  strains,  .         .          Art.  681 

Dimensions  of  parts  suffering  tensile  strains,  Art.  682 

"  "       "  "  compressive  strains,  Arts.  683 

to  687 


DIRECTORY.  573 

FLOOR-ARCHES — TIE-RODS. 

Horizontal  strain,       . (240.) 

Uniformly  distributed  load,  area  of  rod,      .  .         .     (241) 

Load  per  superficial  foot,         "      "     "      ,         .  .          (242.) 

Banks,  assembly  rooms,  etc.,    "      "     "          .  .         .     (243.) 

"         "     Diameter  of  rod,  .         (245.) 

First-class  stores,                                         "     "  .         .     (246.) 

"       "          "        area  of  rod,    ;  (244) 

SHEARING. 

With  compound  load  on  lever,    .  ...       (38.) 

"      load  at  end  of  lever, (39.) 

"     on  beam, (40.) 

Nature  of  the  strain,  .         .        Fig.  30,  Arts.  172,  173,  174 

Web  of  tubular  girder,          .         .         .         .         .         .     (268.) 

PROMISCUOUS. 

Bridle  irons,  for  headers, (28.) 

Bearing  surface  of  beam  on  wall,         ....        (41-) 
Shape  of  beam  and  lever,         .         .      Figs.  31,  43,  44,  Art.  178 
"      depth  at  any  point,  .         .         .  (?4) 

Cross-bridging,  ....  Chap.  XVIII.,  (201.) 
Deflection  illustrated,  •*  Figs.  57  to  64 

Moment  of  inertia  illustrated,  .  .  .  •  Figs.  69  to  72 
Forces  in  equilibrium  illustrated,  .  .  Figs.  81  to  84 

Diagrams  of  forces  illustrated,  .  .  .  Figs.  85  to  88 
Force  diagrams,  .  .  .  Chapters  XXII.  and  XXIII. 
Building  materials,  weights  of  ...  Table  XXII. 


INDEX. 


PAGE 

American  House  Carpenter,  sliding  strains 506 

"          manufacture  of  rolled  iron  -beams,        .     . 313 

"         woods,  constants  (or, • 499 

"        experiments  on 504 

"          wrought-iron,  constant  for, 499 

elasticity  of, 232 

Anderson, experiments  made  by  Major 500 

Angle  irons  in  plate  beam, 312 

Approximate  formulas  discussed, 183 

"              value  of  resistances, 226,  227 

Arch,  area  of  cross-section  of  tie  rod  of  floor 347 

Arches  and  concrete  floors,  weights  of,        340 

"        for  floors,  general  considerations, 345 

tie -rods  for  brick 346 

"        where  to  place  tie-rods  in  brick 348 

Arched  girder  of  cast-iron,  and  tie-rod 396 

"       substitute  for  iron, 398 

Architect,  his  liability  to  err 28 

tables  save  time  of  the, 495 

too  busy  to  compute  by  rules, 495 

Architect's  knowledge  of  construction 27 

Area  of  cross-section,  resistance, 31 

"         of  tie-rod  of  floor  arch, 347 

Arithmetical  computation  of  strains  in  truss, 486 

progression,  the  sum, 147,  148 

series , 151 

"       coefficients  form  an, 226 

Arithmetically  computed  strains 168 

Ash,  resistance  of,       ...                    120 


INDEX.  575 

PAGE 

Assembly  halls,  formula  for  solid  floors  of, 502 

rolled-iron  beams  for,        498 

rooms,  strains  in,  the  same  as  in  dwellings, 88 

"       and  banks,  load  on  floors  of, 340,  341 

"  "       load  on  floors  of, 88 

"  "          "      "    tubular  girders  for, 380 

"  "       rolled-iron  beams  for, 495 

"       "         "         "  Table  XVI1L, 526,  527 

"       "     carriage  beams  with  two  headers  and  one  set 

of  tail  beams,  for 358 

"       rolled  iron  carriage  beams  with  two  headers  and  two  sets 

of  tail  beams,  for, 354,  356 

"       rolled-iron  carriage  beams  with  three  headers,  for,     360,  362 

"       rolled-iron  headers  for  floors  of, 349 

"       rule  for  floors  in, 261 

"         "      "    tubular  girders  for, 380 

tie-rods  for  floor  arches  of, 347 

Auxiliary  formula  for  carriage  beams, 193 

Baker,  Strength  of  Beams,  Columns  and  Arches, 446 

"  "          "         "  "  "         "       ratio  by,     ......     382 

'*       formula  for  posts, 446 

"       on  compression  of  materials, 446 

Banks,  formula  for  solid  floors  of, 502 

"       load  on  tubular  girder  for 380 

"       rolled-iron  beams  for, 495,  498 

"    Table  XVIII. , 526,  527 

"         "      carriage   beams  with  two  headers   and   one  set  of  tail 

beams,  for, 358 

"         "      carriage  beams  with  two  headers  and    two  sets   of  tail 

beams,  for, 354,  356 

"      carriage  beams  with  three  headers,  for 360,  362 

"      headers  for  floors  of, 349 

"       tie-rods  for  floor  arches  of, 347 

"       rule  for  tubular  girders  for, 380 

load  on  floors  of, 340,  341 

Barlow's  constants  for  use  in  the  rules, 499 

"        experiments  on  woods, 233 

"        expression  for  elasticity, 232 

Bays  in  a  framed  truss,  number  of, 426,  428 

Beam  and  lever  compared, 244 


576  INDEX. 

PAGE 

Beam  and  lever  compared,  deflection  in, 237 

"         "         "    their  symbols  compared, 49 

"      device  for  increasing  the  strength  of,     ...           402 

"      distributed  load  on  rolled-iron 337 

"      ends  shaped  to  fit  bearings, 122 

"      load  for  a  given  deflection  in  a, 245 

"      of  economic  form,       163 

"       "  equal  strength, 163 

"      rules  for  dimensions  of  deflected 248 

"      shaped  as  a  parabola, 124 

"      values  of     U,    /,    b,    d    and     &     in  a, 253 

"       "      W,   /,    b,   d      "        6     "  "    . 248 

"      with  load  distributed,  rules  for  size  of, 253 

Beams,  formula  for  deflection  of,  . 229 

"        general  rule  for  strength  of, 92 

"        of  dwellings,  general  rule  for  strength  of, 89 

"         "  wood,  their  weight 79 

"  .     "  warehouses  to  resist  rupture, 260 

"        comparison  of  rolled-iron,  plate  and  tubular,        367 

"        should  not  only  be,  but  also  appear  safe, 211 

strains  in,  graphically  expressed, 177 

Bearing  surface, 122 

"         of  beams  on  walls,         121 

Bearings,  beams  shaped  to  fit, 122 

Bending,  a  beam  is  to  resist 211 

and  appearing  dangerous,  beam  safe,  yet 235 

"          in  good  floors  far  within  the  elastic  limit, 239,  243 

its  effects  on  the  fibres, 35 

"          moment  of  inertia,  resistance  to, 314 

rafter  to  be  protected  from, 479 

"          resistance  to 221 

Bent  lever,  equilibrium  in, 42 

Bow-string  iron  girder, 396 

"         "         "         "        substitute  for, 398 

"         "         "         "        unworthy  of  confidence, 396 

Bow,  Economics  of  Construction, 402,418,425 

"     has  written  on  roofs, 459 

Braces  in  truss,  dimensions  of, 490,  491 

Breaking  and  safe  loads  compared, 68 

"         load  of  unit  of  material, 69 


INDEX.  577 

PAGE 

Breaking  load,  the  portion  to  be  trusted, 69 

'*         weight, 267. 

"  "          compared  with  safe  weight, 235 

"  "          index  of, 51 

"  "          per  inch  sectional  area,  tensile,  Table  XLIV.,       .     .     .      563 

"  "  "      "      surface,  sliding,  Table  XLV.,  ......     564 

"  "  "    unit  of  material,  transverse,  Table  XLII.,        .     .     .     561 

Breadth  from  given  depth  and  distance  from  centres 92 

"        in  first-class  stores, 265,  266 

"        its  relation  to  depth, 33 

"        of  beam,  rule  for 248,  249 

"         "        "      in  dwellings, 262,  263 

"         "        "      with  distributed  load,  rule  for, 254 

"         "   header,  rule  for, 271 

"   lever          "      "         .     .     . 250,  256,  257 

"        proportioned  to  depth,  rule  for 73 

Brick  arch  a  substitute  for  iron  arch  girder 398 

"         "     for  floor,  rate  of  rise, 346 

"         "     less  costly  than  cast-iron  arch, 399 

"      arches  and  concrete  filling, 345 

"       for  floors,  general  considerations 345 

"          "       tie-rods  for, 346 

"       where  to  place  tie  rods  in, 348 

Bridge,  greatest  load  on,  80 

Bridges,  Conway  and  Menai  Straits  tubular, 367,  368,  378 

Bridged  beam,  resistance  of  a, 304 

Bridging  causes  lateral  thrust, 303 

"          for  concentrated  loads, 88 

"          floor  beams 302 

"          in  floors  tested, ...     303 

"          increased  resistance  due  to,     .     - ....     310 

"          measure  of  resistance  of, 304 

"          number  of  beams  resisting  by,     ...          .  • 309 

principles  of  resistance  by,       .  304 

"          useful  to  sustain  concentrated  loads,  .     .          309 

Bridle  iron  and  carriage  beam 98,  195 

"         "     load  upon  a, * 98 

"         "     rule  for  a, 98,  99 

"         "     to  be  broad, 99 

Britannia  and  Conway  tubular  bridges 328,  368 


5/8  INDEX.         * 

PAGE 

Buckling  or  contortion  of  a  tubular  girder 377 

Building  materials,  weights  of,  Table  XXII., 533,534,535 

Buildings  require  stability,        27 

"          requisites  for  stability  in, 28 

Burbach,  large  rolled-iron  beams  from 313 

Buttresses  to  support  roofs  without  ties 459 

Calculus  and  arithmetic  compared,          318,  320,  321 

"    scale  of  strains 161 

"        applied,  result  by  the , 323,  325 

"        coefficient  defined  by  the 227 

"        strain  by  distributed  load, 157 

"  "      defined  by  differential 180,  18.;. 

"        strains  in  lever  by  differential 168 

Cape's  Mathematics,  forces  shown  in, • 404 

references  to, 160,  164,  169 

Carriage  beam  and  bridle  irons 98 

"  "        "     headers 94 

"      auxiliary  formula, 193 

"      definition, 95 

"      for  dwellings,  precise  rule, 273,  285 

"        "   first-class  sto.res,  precise  rule, 275,  282,  286 

"     '  formula  not  accurate, 183 

"       load  on  a, 98,  107 

"       of  equal  cross-section, 103 

"       precise  rule,     h     greater  than     n, 281 

"  "  "  "         h     less  "        n 280 

"       special  ruleSj 281 

"       with  one  header,  rule, 99 

"  "         ."  .    "         "         rolled-iron 351 

"       "         "         for  assembly  rooms,  rolled-iron      .     .     .     351 

"    banks,  rolled-iron 351 

"  "  "       "         "          "    dwellings 272 

"     .      "  "  "  rolled-iron 351 

"    first-class  stores, 273 

"       rolled-iron        .     .     .     352 

"    two  headers, 101 

"       "     -    "          for  dwellings,  precise  rule, 292 

"         4<  "    first-class  stores,  precise  rule,        .     .     292 

"       "   equidistant  headers,  precise  rule,        287 

"          for  dwellings,  precise  rule,       289 


INDEX.  579 

PAGE 

Carriage  beam,  with  two  equidistant  headers,  for  first-class  stores,  precise 

rule 289 

"   headers  and  one  set  of  tail  beams, 106 

*•           "          «•       "         'I           "       "     "    "    "         "      precise  rule,    .  283 
"           "           "       "    equidistant  headers  and  one  set  of   tail   beams, 

precise  rule, 290 

*«           "          "       "   headers  and  one  set  of  tail  beams,  for  dwellings,  277 

rolled-iron 358 

"  "  "  "  headers  and  one  set  of  tail  beams,  for  first-class 

stores 278 

"  "  "  "  headers  and  one  set  of  tail  beams,  for  first-class 

stores,  rolled-iron 359 

"  "  "  headers  and  two  sets  of  tail  beams,  .  104,  192,  194 

«c  it  «  •»  .«  .»  "  «»  ...  «•  »  precise  rule,  279 

"  "  "  "  '•  "  rolled-iron  .  353 

)  275 


precise  rule, 282 

"  "  "       "    headers  and  two  sets  of  tail  beams,  for  dwellings, 

rolled-iron 354,  356 

"  "  "       "    headers  and  two  sets  of  tail   beams,  for  first-class 

stores .  276 

"  "  "       "    headers  and  two  sets  of  tail  beams,  for  first-class 

stores,  rolled-iron 355,  357 

"           "          "    three  headers 195,  196,  197 

"           "                                            for  dwellings,  rolled  iron    .      ,     .       360,  362 
"           "           "         "           "           "  first  class  stores,  rolled-iron          361,  364 

"           "           "         "                      the  greatest  strain  at  middle  header,     .  297 

"           "                                                                     "       "  outside      "           .  294 
"           "           '                      "                                      "        "  middle       >; 

for  dwellings 298 

"  "          "        "   headers,  the  greatest  strain  at   outside  header, 

for  dwellings, 295 

"  "          •'        "   headers,  the  greatest  strain  at  middle  header,  for 

first-class  stores, 299 

"  "          "        "   headers,  the  greatest  strain  at  outside  header,  for 

first-class  stores, 295 

"          "        "   headers  and  two  sets  of  tail  beams,    .     .     .       200,207 

Cast-iron,  compression  and  tension  in, 387 


S8O  INDEX. 

PAGE 

Cast  iron  resists  compression  more  than  tension, 45 

44      "      superseded  by  wrought-iron, 386 

beam,  load  at  middle,  Hodgkinson, 383 

44      "      arched  girder  with  tie-rod, 396 

"      "          •'•  4<       tie-rod  for, 396 

"      "          4'  4t       form  of  web  of, 392 

"      "  •'  "       for  brick  wall  with  three  windows,        .....     394 

*4      4<  4<  4'       load  at  any  point  of,  rupture, 390 

•  ••-          "         "          middle  of, 389 

"  "       proportion  of  flanges  of, 386 

44      "  44  "       safe  distributed  load,  effect  at  any  point  on,     .     .     391 

"      "          "  "       safe  load  at  any  point  on 391 

44      44          '4  "       two  concentrated  weights  on 392 

44      "      girders,  chapter  on, 386 

Ceiling  of  room  plastered, 303 

44       to  be  carried  by  roof  truss,  weight  of, 481,  483 

44       weight  of,    .......'-. 78 

Cement  grout  for  brick  arches  of  floors, 345 

Centennial  Exposition,  rolled-iron  beams  at, 313 

Centre  of  gravity,  load  concentrated  at  the 60 

Centres,  distance  from, 91 

Cherry,  resistance  of,        .     .     .     .  • 120 

Chestnut,       4f          " 120 

Chord,  framed  girder  with  loads  on  each 433 

44       of  framed  girder,  allowance  for  joints,  etc.,  in, 445 

area  of  uncut  part  of, 444 

44        strains  in  lower        439 

44        "  upper        440 

41       and  struts  of  framed  girder,  upper 448 

"         compression  in  upper        445 

Chords  and  diagonals,  gradation  of  strains  in 432,  435 

of  framed  girders  usually  of  wood 444 

Civil  Engineer  and  Architects'  Journal, 82 

Clark,  moment  of  inertia,  by  Edwin 328 

Clark's  formula  only  an  approximation 328 

useful  in  certain  cases, 330 

Clay  has  but  little  elasticity, 211 

Coefficient  of  strength  for  tubular  girder, 368 

Coefficients  in  rule  for  floors  of  dwellings, 261,  262 

44  4t     4<      "       4t       "  first-class  stores 264,  265 


INDEX.  581 

PAGE 

Components  of  load  on  floor, 339 

Compound  load,  assigning  the  symbols, 187 

44  "       dimensions  of  beam, 187 

"  "       general  rule,        199 

44  "       greatest  strain  from, T82 

44  "       maximum  moment, 188 

"       strain  analyzed, 178 

44  "       strains  and  sizes, 171 

44       on  floors,        ,     .     339 

"  "         "  lever,  the  effect  of, 171,  174 

strains  graphically  expressed, 177 

Compressibility  of  fibres 37 

Compression  balances  extension, 45 

dimensions  of  parts  subject  to 490 

graphically  shown, 115 

resistance  to, 45 

"  and  extension  of  fibres,  strength, 35 

"  "  "          summed  up, 228 

44  4'    tension,  fibres  resisting, 403 

"  44         "         of  cast-iron, 387 

"  "         44         rupture  by, 313 

of  fibres  at  top  of  beam, 42 

44  44  struts,  rule  for, .       447,  449 

44       "        application  of  rule, 447 

in  struts  and  chord  of  framed  girder, 445 

Rankine,  Baker  and  Francis  on,   . 446 

44  Tredgold  and  Hodgkinson  on, 446 

Compressive  and  tensile  strains, 408 

strain  in  rafter  increased, 474 

Computation  by  logarithms,  example  of, 311 

4'  of  moment  of  inertia, 315 

strains  in  framed  truss  by, 416 

44  to  check  graphic  strains, 132 

Concave  side  of  beam,  fibres  compressed  at, 37,  45 

Concentrated  and  half  of  distributed  load  equal  in  effect  at  any  point,  .     .     162 

"          load,  bridging  useful  in  sustaining, 302,  309 

"  "      resistance  of  bridging  to  a, 308 

"  "      location  of  greatest  strain 181 

"  loads,  a  series  of, 155 

"  "  approximates  a  distributed  load,       .     .     .     155 


5cS2  INDEX. 

PAGE 

Concentrated  and  distributed  loads, 155,179,181,182,183,274 

••  "  "  "      compared, 61,  62,  63,  161 

"  "  ••  "      graphically  expressed, 177 

"  "  "  "      on  beam, 252 

•«  "  "  "      size  of  beam, 182,191 

"      on  lever, 171,  174,  255 

loads,  a  distributed  and  two 184,187,191 

"  "         "       "  "     three       .     .     .      195,  197,  198,  199,  203,  292 

"  "        middle  load  of  distributed  and  three 296 

Concrete  and  brick  arches,  weight  of, 340 

"         filling  over  floor  arches, 345 

Conflagrations  resisted  by  solid  floors, 500 

Constant     F    for  deflection,  values  of,  Table  XLIII., 562 

Constants  for  tubular  girders. 369 

"  "    use  in  the  rules,  .     .      * 499 

Table  XX 530,  53i 

"          from  experiments  in  certain  cases, 244 

"          how  derived 505 

precautions  in  regard  to, 243 

Converging  forces  readily  determined 418 

Convex  side  of  beam,  fibres  extended  at 37,  45 

Conway  and  Britannia  tubular  bridges, 328,  367,  368 

Construction  defined, 27 

Tredgold  an  authority  on, 81 

"  weight  of  the  materials  of, 78,261,264 

weights  of  materials  of,  Table  XXII., 533,  534,  535 

•    "  in  a  roof,  weight  of  the  materials  of, 480,  483 

Cross-bridging, 303 

"  "         assistance  derived  from, 308 

"  "         dowels  act  as, 501 

Cross-furring,      . 303 

Cross-section,  moment  of  inertia  proportioned  to, 314 

Crush,  bricks  liable  to 346 

Crushing  strains,  tests  of  woods  by, 506 

"  "         in  Georgia  pine,  locust  and  white  oak,  Table  XL.,      .     .     559 

"  "         "  spruce,  white  pine  and  hemlock,  Table  XLL,     .     .     .     560 

"         weights  per  inch  sectional  area,  Table  XLVI 565 

Curve  and  tangent,  point  of  contact  defined, 179,  184 

"       of  equilibrium  is  a  parabola, 416 

"       "  "  stable  and  unstable 416 


INDEX.  583 

PAGE 

Dangerous,  beam  though  safe  may  bend  and  appear 235 

Deflected  lever,  rules  for  size  of,  . , 256 

Deflecting  energy, 211 

"         of  weight  on  lever, 229 

energies  in  beam  and  lever,  ....  * 251 

power  of  concentrated  and  distributed  loads, 252 

Deflection  and  rupture  compared, 211 

excessive  under  rules  for  strength, 77 

resistance  to, 221 

by  distributed  load,  rule  for, 255 

directly  as  the  weight, 304 

"         "    "  extension, 214,215 

"  "   .      "     "  length,  .     . 217,218 

"  "          "     "  force  and  length, 216 

total,  directly  as  the  cube  of  the  length, 218 

"         "     "    weight  and  cube  of  length, 219 

values  of  constant     F,    Table  XLIII 562 

of  beam,  effect  at  bearing 121 

"       "     formula  for, 229 

"       "      load  forgiven, 245 

with  load  at  middle, 242 

in  floors,  rate  of, 240 

not  to  be  excessive, 211 

to  the  limit  of  elasticity, 237,243,246,247 

within  elastic  limit, 245 

of  beams  not  to  be  perceptible, 260 

per  lineal  foot,  rate  of, 239,  261,  264,  267,  342 

"          injurious  to  plastering,  perceptible 260 

in  good  floors  far  within  the  elastic  limit, 239,  243 

'•          of  beam  with  distributed  load, 251 

rule  for  dimensions  of  beam 248 

of  bridged  beams  tested, 303 

"  rolled-iron  beams,  load  at  middle, 331,  332 

"  "  lever  and  beam  compared 237 

"      amount  of,  ...     t 213 

"      test  of, 245 

"     load  for  a  given, 247 

"  "      "     by  distributed  load, •  .     255 

"      "      to  limit  of  elasticity, 247 

"  "      "      rule  for, 229,  244,  256,  258 


584  INDEX. 

PAGE 

Peflection,  dimensions  of  lever, 251 

"       4t      rules  for, 250 

is  as  the  leverage,  the  power  to  resist 223 

Demonstration  of  scale  of  strains 134 

Depth,  its  value,  test  by  experience, 33 

"       and  length,  ratio  between, 240 

"       relation  to  weight  and  fibres 36 

"        in  proportion  to  weight, 44 

"        denned  for  compound  load, 178 

"        relation  to  breadth, 33 

"       proportional  to  breadth,  rule, 73 

"       from  given  breadth  and  distance  from  centres, 92 

"       of  simple  Jaeams  necessarily  small, 402 

in  a  beam,  the  importance  of, 312 

of  beam  proportioned  to  load,  square  of, 123 

"        "      "      rule  for, 248,  249 

"      with  load  distributed,  rule  for, 254 

"        "      "      in  dwellings, 262,  263 

"       "       "  first-class  stores, 265,  266 

"  lever,  uniform  load, 169 

"        "       "      promiscuous  load 175 

"        "      "      rule  for, 251,  256,  257 

"        "  framed  girder,  rule  for, 424 

"       in       "  "         objectionable, 422 

"       and  length  of  framed  girder 422 

"       to  length  in  tubular  girders,  ratio  of, 382 

Depths  analytically  defined,  varying 137 

"        expressions  for  varying 141 

"       demonstrated,  rule  for  varying, 135 

"       compound  load  on  lever,  scale  of, 172,  173 

Design  for  a  roof  truss,  selecting  a 481 

Destructive  energy, ,     .  47, 48,  116,  121,  151 

"         its  measure, 53 

"         symbol  of  safety, 71 

"         and  resistance, 53 

"         load  at  any  point, 57,  129 

"        from  two  weights, 133 

"      several  weights, 62, 66,  67 

"         on  lever, in 

power  of  weight  and  resistance  of  material, 68 


INDEX.  585 

PAGE 

Diagonals,  gradation  of  strains  in  chords  and, 432,  435 

"  of  framed  girder,  strains  in  the, 436 

top  chord  and, 448 

Diagram  of  forces  described 418,419,421 

"         "       "       order  of  development  of, 421 

"         "       "       gradation  of  strains  in, 433 

"         "       "       in  fra'med  girder 429 

Diagrams  and  frames,  reciprocal, 418 

"         correspondence  of  lines  in  frames  and,    .........     462 

Differential  calculus, 158 

"         computation  by, 228 

"  "         strain  denned  by .       180,184 

"  "         strains  in  lever 168 

of  variable,  moment  of  inertia, 318,319,  322 

Digest  or  directory  of  this  work, 566 

Dimensions  of  beam  for  compound  load, 182,187,189 

"       4t      at  given  point,  for  compound  load, .     .......      189 

"      "     load  at  any  point, 129 

"       "      when     h     equals     «, 190 

"         "         h     exceeds     n, 191 

Directory  or  digest  of  this  work, ,     .  566 

Distance  from  centres  of  beams 80,  91 

"   girders, ...       94 

"         "    rolled-iron  beams, 341,  342,  498 

in  dwellings, 262 

rolled-iron  beams, 343 

"    first-class  stores, 265 

"        rolled-iron  beams,       ....     344 

Distributed  load,  strain  by  the  calculus, 157 

"       effect  at  any  point, .      161 

"  "       equal  in  effect  at  any  point,  concentrated  and  half  of,      .      162 

"  "       on  floors '77 

"  "        "   beam, 75 

"  "       deflection  of  beam  under, 251 

"  "       shape  of  side  of  beam, ••  .     .     .     .     .     .     162 

"  "       on  lever, 74 

"       deflection  of  lever  by,     , 255 

"  "       shape  of  side  of  lever, 170 

on  rolled-iron  beam, 337 

"  "       "   cast  iron  girder,     .     .     . 389 


586  INDEX. 

PAGE 

Distributed  load  on  tubular  girder, 368 

««             »«        "         "           "       size  at  any  point, 372 

"    .       and  concentrated  loads  compared 61,62,63,155,161 

"             "             "                 ''        on  beam  compared, 252 

"             "             "                 "        "    lever         "             255 

•'             "     one  concentrated  load, .     179,  182,  183,  274 

"     graphic  representation,      ....  177 

"     size  of  beam 182 

"                                                        "     on  lever, 171,  174 

"     two                            loads,   . 184,  187,  1 91,  195 

size, 191 

"     three                                       ....    195,  197,  198,  199,  203,  292 

"        middle  load, 296 

Drury,  testimony  on  loading, 82 

Dwellings,  load  on  floors  of, - .,88,  340,  341 

"            floor  beams  and  headers  for 495 

"            rule  for  floors  in 261 

"              "       "    headers  in 271 

"           values  of    c,    /.    b   and   d  in  floors  of,        262 

"            rule  for  solid  floors  of, 502 

.    "           hemlock  beams,  Table  1 508 

headers,  Table  IX 516 

"            Georgia  pine  beams,  Table  IV., 511 

"                 "           "      headers,  Table  XII., 519 

"           spruce  beams,  Table  III 510 

headers.  Table  XL, 518 

"            white  pine  beams,  Table  II 509 

"               "         "      headers,  Table  X., 517 

carriage  beam,  precise  rule, 282 

"         with  one  header 272 

"                 "            "           "      two  headers,  precise  rule,  ......  292 

"            "           "        "    equidistant  headers,  precise  rule,  .     .  289 

"        "    headers  and  one  set  of  tail  beams.       .  277 

"        "         "          "     two  sets      "         "             .  275 
three    "         the  greatest  strain  at  middle 

header,     .          298 

"         "        "         the  greatest  strain  at  outside 

header, 295 

rolled-iron  beams  for,      . 498 

"    Table  XVIII.,  526,  527 


INDEX.  587 

PAGE 

Dwellings,  rolled-iron  beams  for,  distance  from  centres, 343 

tie-rods  for  floor  arches  of 347 

"           rolled-iron  headers  for, 349 

"  "         "     carriage  beams  with  two  headers  and  one  set  of  tail 

beams, 358 

«V  "         "  "  "         "         "    headers    and    two   sets    of 

tail  beams,  .     .     .       354,  356 
"  <4         "  ••  "         "      three  headers,     ....       360,  362 

"           load  on  tubular  girders  for 380 

rule  for                     "         "          380 

Economical  depth  of  framed  girder, 425 

"  tubular     '.'        382 

"             form  of  beam,        •  .     .  163 

"               "       "  rolled-iron  floor  beam 312 

"               "       "  roof  truss,  more, 469 

Elastic  curve  defined  by  writers 213 

"       limit,  bending  in  good  floors  far  within  the, 239,  243 

"          "       fibres  strained  beyond  the, 235 

"          "        important  to  know  the, 212 

44          "        in  elongation  of  fibres, 236 

"          "        symbol  for  safety  at  the, 239 

"        power  of  material,  knowledge  of, 244 

"        substance  in  soles  of  feet, 84 

Elasticity  are  exceeded,  rupture  when  limits  of 212 

defined,  limits  of,                      212 

"          for  wrought-iron,  modulus  of, 232 

"          of  floor,  moving  bodies, 84 

"          possessed  by  all  materials, 211 

Elements  of  rolled-iron  beams,  Table  XVII., 524,  525 

Elliptic  curve  for  side  of  beam, 164 

Elongation  of  fibres 214 

"           "       "       graphically  shown, 236 

English  rolled-iron  beams,  large, 313 

44         wrought  iron,  elasticity  of, ,  232 

Equal  weights  equally  disposed, 141.  143,  144,  146 

4<           "          general  results, 146 

"           "          strain  at  first  weight, 147 

44           "             4<       "  second  weight 148 

44           44             4t       "  any  weight, 150 

Equally  distributed  safe  load,  rule  for,       70 


588  INDEX. 

PAGE 

Equation,  management  of  an, 71 

to  a  straight  line, 171 

Equilibrated  truss,  strains  in  an 408 

Equilibrium  at  point  of  rupture,   .           68 

"            measure  of  forces  in 407 

Equilibrium  of  pressures , 38 

"          "    resistances  of  fibres 45 

"          stable  and  unstable 416 

"          three  forces  in 406 

Error  in  rules  on  safe  side,        ... 183 

Euclid's  proposition  in  a  triangle, 486 

Excess  of  material  by  rule  for  carriage  beam, 183 

Experiment  as  to  action  on  fibres 36 

"  on  India-rubber, ; 212,  213 

"              "    New  England  fir. 233 

"              "    white  pine  units, 32 

Experiments  by  transverse  strain, 504 

on  American  woods, 504 

"    cast-iron,  Hodgkinson, 386 

"    model  iron  tubes 369 

"    side  pressure .  120 

."            "    tensile  and  sliding  strains 505 

"    timber,        30 

"    units,  conditions 32 

."            "    weights  of  men,  . 85 

"             "    woods,  by  crushing 506 

"             "    wrought-iron, 232 

"            rules  useful  in,    .     . 68 

Experimental  test  of  cross-bridging, 303 

Extension  and  compression  of  fibres,  strength, 35 

"           "                              summed  up, 228 

as  the  number  of  fibres,  resistance  to 222 

balances  compression, •     .     .  45 

directly  as  the  area  and  depth,        222 

.  "      force 212 

"      length 213 

graphically  shown,  resistance  to 221 

measured  by  reaction  of  fibres, 222 

of  fibres 37 

"       "     at  bottom  of  beam, 42 


INDEX.  589 

PAGE 

Extension,  resistance  to, 45 

Factory  floors,  load  on, 78,  79 

Fairbairn's  experiments, 500 

Fairbanks  Scale  Co.,  testing  machine  by 504 

Falling  body,  the  force  exerted  by  a, 84 

Feet,  elastic  substance  in  soles  of,      .     .     , 84 

Females,  weights  of, 83 

Fibres,  crushed  on  wall, 121 

"       crushing  in  direction -of, 506 

"       elongated  to  elastic  limit, 236 

"       end  and  side  pressure  on, 120 

"       extended  or  compressed 212,  214 

"       extension  of,  graphically  shown, *     ...  222 

"       in  a  tie-beam,  consideration  of, .     i     .  488 

"       load  should  not  injure  the, 69 

"       measuring  extension  of  the, , 221 

"       power  of  resistance  as  the  depth,   , 46 

resistance  as  the  depth  of  beam, 43 

"               "            "     "  leverage, 223 

"              "           directly  as  the  depth, 36 

"              "           to  change  of  length, 46 

"               "            "  extension, 222 

"           "  horizontal  strain,        43 

"                           "  side  pressure, 120 

"       resisting  compression  and  tension,     ,     .          403 

"       strained  beyond  elastic  limit, 235 

"       strength  due  to  their  coherence, 35 

Fire,  resisted  by  solid  timber  floors 500 

"     wooden  beams  liable  to  destruction  by, 312 

Fireplaces,  framing  for 95 

First-class  stores,  carriage  beams  with  one  header, 273 

•    "    .    "         "        floor  beams, 264 

"      and  headers, 495 

"        "         "        rule  for  headers, 27l 

"        "         "        formula  for  solid  floors, 5C-3 

"         *•        load  on  floors, 339>  34 1 

"        "         "          "      "    tubular  girders, 380 

"        "         "        rule  for  tubular  girders, 381 

"        *•         «'        rolled  iron  beams, 495 

"       "         "             "        "        "       distance  from  centres 344 


590 


INDEX. 


PAGE 

First-class  stores,  rolled  iron  headers,        350 

•«        "      •  "             "        "     carriage  beams  with  one  header,       ....  352 
«•        "         <v             "        "                                    "    two  headers  and  one  set 

of  tail  beams,  .     .     .  359 
"        "         "                                                                     two  headers  and  two  sets 

of  tail  beams,  .     .     .  357 

"        "         "             "        "          "             "          "     three  headers,  .     .     361,  364 

"        "        "        tie-rods  in  floor  arches,        348 

"        "         "        values  of    c,   /,    b   and    dy       .     : 265 

Five  equal  weights,  graphic  strains 144 

Flanges  an  element  of  strength, 313 

and  web,  proportions  between,      .     .     .- 313 

"           "       "      in  cast-iron  girders,  relation  of, 387 

"           "       "      moment  of  inertia  for,    . 327 

in  cast-iron  girders,  proportion  of, 387 

'         equal,  top  and  bottom 314 

"        of  cast-iron  girders,  proportion  of, 386 

"         "  tubular  girders,  construction  of 374 

"        equal,  top  and  bottom        371 

tension  in  lower, 370 

"        for  floors,  area  of. k     .  377 

minimum  area  of, 383 

"         "            "        thickness  of,    .          373 

to  predominate  over  the  web,       , 314 

Flexure  and  rupture  compared, 267 

"        rules  compared 235,  293 

weights  producing  compared, 237 

floor  beams  by  rules  based  on,       .     .     .     , 77 

formula  for  denned, 230 

"        moment  of  inertia,  resistance  to 3r4 

of  floor  beams,  resistance  to. 260 

"        resistance  to, : 221 

"   rules  for, 242 

value  of    F%    the  symbol  of  resistance  to, 230 

Floors,  application  of  rules  for  strength  of, 77 

load  on  rolled  iron  beam 34° 

not  always  strong •     •  29 

of  solid  timber,  Table  XXI 532 

1    warehouses,  factories  and  mills,     , 7& 

per  superficial  foot,  load  on, 2^r 


INDEX. 


591 


PAGB 

Floors,  safe, 28 

"        severest  tests  on, 85 

"        strength  of, 29 

"  beams  in,  rule  for,  .     .     , .  77 

"              "         "  test  by  specimens, 29 

"       tubular  girders  for,  rule  for,        377 

"        weights  of,  in  dwellings, 80 

Floor  arches,  general  considerations, 345 

"         of  parabolic  curve, 346 

"          "         tie-rods  for, .     .     .     ^    .     .  346 

"         "              "         "   area  of  cross-section  of, 347 

"         "              "         "  where  to  place 348 

"      beams,  general  rule  for,       , 89,  92 

"          "        bridged 302 

"    "   nature  of  load  on, 78 

"   load  on,  rule  for, 78 

"          "        of  dwellings,  modified  rule  for, 2bi 

"        resistance  to  flexure  of, 260 

"          "        stiffened  by  bridging, 310 

"          "        stiffness  of,  rule  for 260 

"        of  wood,  Tables  of, 496 

"        "       "       and  iron,  Tables  of, 495 

"          "        "  iron,  distance  from  centres,        341 

"          "      Georgia  pine,  for  dwellings,  Table  IV 511 

"             "         "        •'  first-class  stores,  Table  VIII., 515 

"          "       hemlock,  for  dwellings,  Table  I., 508 

"         '«                          "     first-class  stores,  Table  V., 512 

"         "       spruce,       "     dwellings,  Table  III 510 

"            "             "     first-class  stores,  Table  VII., 514 

"         "        white  pine,  for  dwellings,  Table  II.,          509 

"         "            "        "       "    first-class  stores,  Table  VI., •   .  513 

bridging  tested, 303 

"         "        openings,  carriage  beams, 195,  196 

"         "        planks,  their  weight, 79 

Force  exerted  by  a  falling  body, 84 

and  frame  diagrams  correspond,  lines  of, 462 

Forces  and  lines  in  proportion, 405 

described,  diagram  of, 418,  419,  421 

"       in  a  framed  girder, 428 

"        "  "  truss,  graphically  shown, 417 


592  INDEX. 

PAGE 

Forces  shown  by  a  closed  polygon, 418 

Force- diagram,  example  of  constructing  a,      ?.;..., 483 

"      =  for  a  roof  truss, .   .     .     .     .  461,  462,  463,  465,  466,  468,  469,  472 

"          .  **«        form  a  closed  polygon,  lines  in  a, 485 

"         line  of  weights  for  a, 464,  466 

"             "         of  a  roof,  measuring  the, 485 

"    •        "         of  an  unsymmetrically  loaded  girder, 455 

"             "         scale  of  weights  in  a, 483 

"      diagrams,  strains  in  trusses  compared  by, 469 

Form  of  beam  for  distributed  load, 162 

"     *"    lever  "                            " 171 

'"  "     "      "    compound      " 173 

"      "   iron  beam,  economical 312 

Formula,  comparison  of    F    with     E      of  common,  .     , 232 

"         for  resistance  to  flexure, , 232 

"   solid  floors, , 502 

'•       "         "-    reduction, 501 

"         management  of  a, 89,90 

"         practical  application, , 71 

Four  equal  weights,  graphic  strains, 143 

Frames  and  diagrams,  reciprocal, •  418 

Framed  girder,  allowance  for  joints,  etc.,  in  chord, 445 

"           "         area  of  imcut  part  of  chord, 444 

"         bearings  of  metal  for  struts, 450 

"         compression  in  chord  and  struts, 445 

"         compromise  of  objections, 423 

cost  inversely  as  the  depth,    ...     , 423 

"         diagram  offerees  in, 429 

"         economical  depth, 425 

'*        forces  in, 428 

"         horizontal  thrust  in, 403 

'*        irregularly  loaded, 451 

'*         its  relation  to  a  beam 402 

liable  to  sag  from  shrinkage, 450 

"         minimum  of  strains  in, , 426 

"         number  of  bays  or  panels, 425,428 

"         peculiarity  in  strains  of, 432 

"         proportions  of,      .- 422 

"         resistance  to  tension  in, 443 

"         rule  for  depth, 424 

"           "         series  of  triangles  in, .  425 


.  INDEX.  593 

PAGE 

Framed  girder,  strains  in  diagonals  of, 436 

44           "              "       "  lower  chord  of, 439 

"       "  upper      "       "         440 

"           "         system  of  trussing  in,    .          .     t 425 

41           "         top  chord  arid  diagonals  of, 448 

tracing  the  strains  in, 437 

trussing  in,  .     .     .     .     .     .    «' 417,425 

unequal  reactions  of  supports  of, 451 

with  loads  on  each  chord,  . 433 

wrought-iron  ties, in 443 

"       girders,  chapter  on 402 

compression  in,  rule  for, 447 

"          usually  of  wood,  chords  of 444 

truss,  reaction  of  supports  of, 415 

France,  testing  bridges  in 82 

Francis  on  compression  of  materials 446 

Funicular  or  string  polygon, 408 

Furniture  reduces  standing  room, 88 

Galileo's  theory  of  the  transverse  strain, 36 

Geometrical  approximation  to  moment  of  inertia, 315 

series  of  values  of  strains, 476 

Georgia  pine,  resistance  of, 120,  121 

"      beams,  their  weight, 79 

41      floor  beams  and  headers, 495 

"      coefficient  of  in  rule, 261,265 

German  rolled-iron  beams,  large, 313 

Girder  defined,  rule, 94 

"      history  of  tubular  iron, 367 

"      plate  and  jolled-iron,  compared  with  tubular, 367 

Girders,  distance  between, , 94 

"         headers  and  carriage  beams 94 

Graphic  representation  of  strains, 127 

"         strains  checked  by  computations, 132 

"             "        from  two  weights 133 

"     three     "            138 

"             "           "         '4     equal  weights, 141 

"     four       "                      143 

Graphical  representations, ' in 

of  compound  loads, .  177 

"                     "               of  moment  of  inertia, 321 


594  INDEX. 

PAGE 

Graphical  strain  at  any  point,  . '. 127 

"         strains  in  a  beam 114 

"        "  "  double  lever, 113 

Graphically  shown,  horizontal  strains 406 

"         resistance  of  fibres 223 

Gravity,  its  prevalence, 27 

load  concentrated  at  centre  of, 60 

Greatest  load  on  floor, 80 

Hatfield's,  R.  F.,    clock-work  motion, 504 

Headers,  definition, 95 

"        load  upon,   . 96,  196 

"         allowance  for  damage  to. 97 

"         formulas  for 96,  97 

11               "          "     tables  of, 497 

"               '•          "     breadth  of 270 

"        and  trimmers 266 

"         wooden  floor    . 495 

"         rolled-iron  floor 349 

*'         for  dwellings  and  assembly  rooms, 271 

"           "         "                                         "      rolled-iron 349 

"           "  first' class  stores, 271 

"           "     "        "          "       rolled-iron 350 

'*         carriage  beams  and  girders, 94 

".        in  carriage  beam,  two 104 

"         one  set  of  tail  beams  and  two 106 

"         carriage  beam  with  three 200 

"         of  wood,  Tables  of,    .          .          497 

"         Georgia  pine,  for  dwellings,  Table  XII 519 

"             •'           "         "  first  class  stores,  Table  XVI 523 

"         hemlock,  for  dwellings,  Table  IX 516 

"    first-class  stores.  Table  XIII 520 

"•        spruce,  for  dwellings.  Table  XI 518 

"    first  class  stores,  Table  XV 522 

white  pine,  for  dwellings,  Table  X 517 

"    first-class  stores,  Table  XIV., 521 

Hemlock,  coefficient  in  rule  for, 261,  265 

Hemlock,  resistance  of, 120,  121 

beams,  their  weight, 79 

floor  beams  and  headers 495 

Hickory,  resistance  of, 120 


INDEX.  595 

PAGE 

History  of  the  rolled-iron  beam 3T3 

V     "    tubular  iron  girder, 367 

Hodgkinson  on  compression  of  materials 446 

Hodgkinson's  edition  of  Tredgold  on  Cast-iron, -     .  386 

experiments,      . .'....  50° 

"              rule  for  cast-iron,  load  at  middle 388 

"               "set"  in  testing,      ,                 , .     .  5°5 

value  of  elasticity  of  iron, 232 

Hoes'  foundry,  weight  of  men  at 85 

Homologous  triangles,  proportions  by 487 

Hooke's  contribution  to  the  science 37 

Horizontal  and  inclined  ties  compared,  strains  in 472 

strain  in  roof  truss, .    466,473,474 

"              "      resisted  by  iron  clamps 489 

"          strains  in  framed  girders, 439,  440,  441,  443 

"              "         measured  arithmetically,  , 412 

"              "         shown  by  bent  lever, 42 

"              "           "        graphically, 406 

"          thrust  in  a  framed  girder 403 

"          tie,  raise  wall  of  building  to  get 478 

Hypothenuse  of  right-angled  triangle 486 

Important  work  should  be  tested,  materials  in 244 

Inclined  tie-rod  of  truss,  enhanced  strain, 477 

Increased  strains  in  roof  truss  from  inclined  tie, 474,  478 

Index  of  strength  for  unit  of  material.          48 

India-rubber,  experiment  on, 212 

"           "         largely  elastic, 211 

Infantry,  space  required  for,      .,.......' 83 

Infinite  series,  sum  of  an, 476 

"           "        value  of  coefficient, 227 

Infinitesimally  small,  differential  is    ,     .  318,  319 

Insurance  offices,  load  on  tloor  of, 340,  341 

Integral  of  moment  of  inertia,        .     .     .     ,     , 319 

"        calculus,  maximum  ordinate, 180,  184 

Integration,  computation  by, 228 

"             rule  for  strain  in  lever  by 169 

"             strain  by, 159 

Iron  a  substitute  for  wood 312 

"    bolts  and  clamps  for  tie-beam, 489 

"    load  upon  wrought, 99 


'C)6  INDEX. 


Iron  beam,  load  at  middle  upon,        331,  332,  333 

"         "      progressive  development  of, 312 

Jackson's  foundry,  weight  of  men  at, 85 

Kirkaldy's  experiments, 500 

Laminated  and  solid  beams  compared 34 

Lateral  thrust  by  cross-bridging, 303 

Lead  has  but  little  elasticity, 211 

Leibnitz's  theory  of  transverse  strains 37 

Length  and  weight,  relation  between, 65 

"         "    depth,    ratio  240 

"      of  beam,  rule  for 248,  249 

"       '.'      "       with  load  distributed,  rule  for, 253 

"       "      "        in  dwellings,       . 262,  263 

"       "      "        "  first-class  stores,      . 265,266 

"       "  rolled-iron  beam,  load  at  middle, 331,  332 

"       "  lever,  rule  for, 250,  256,  257 

"      and  depth  of  framed  girder, 422 

"       to  depth  in  tubular  girder,  ratio  of, 382 

Lever  and  beam  compared, 244 

deflection  in 237 

"        "        "  "     i      strength     " 55 

symbols    " 49 

"      arms  in. inverse  proportion  as  the  weights, 39 

"      at  limit  of  elasticit)',  load  on 248 

"      by  distributed  load,  deflection  of 255 

deflection  in  a, 213 

destructive  energy  in  a.       , 55,111 

"      dimensions  of  a  deflected, 251 

"      distributed  load  on  rolled-iron, 338 

"      effect  of  weight  at  end  of, 47 

formula  for  deflection  in  a, 229 

"       modified  to  apply  to  a, 54 

"      graphical  strains  in  a  double, 113 

load  at  end  of  rolled-iron, 336 

principle  in  transverse  strains, 38 

demonstration 39 

effect  of  several  weights, 62 

"        unequal  weights, 39 

"      promiscuously  loaded, 175 

depth  of, .     ...     175 


INDEX. 

PAGE 

Lever,  rule  for  deflection  of, 244 

"     "    resistance  of, 48 

"         "     "    strength      " 55 

"      rules  for  dimensions  of  deflected, 250,256 

"      safe  load,  rule, 70 

"    distributed  load,  rule, 70 

"      shape  of  side  of 123 

"      shaped  as  a  parabola, 124 

"      showing  elongation  of  fibres, , 236 

"      strains  like  two  weights  at  ends, 45 

"       measured  by  scale, in 

symbol  showing  strength  of, 47 

"      test  of  deflection  in  a, 245 

"      to  compression,  resistance  of  fibres  of .  228 

44       "  extension                                                       .     .           228 

."       "  limit  of  elasticity,  deflection  of, 247 

"      uniformly  loaded,  strains  in, 168 

4'      values  of  P,    n,    b,    d  and    d   in  a, 250 

"      "     U,    n,    b,    d    "     6    "  '• 256 

41      effect  of  weight  at  end  of, ,     .     .  58 

."      with  compound  load,  strain  in  and  size  of, .  171 

"         "    distributed  load,  the  form  a  triangle, 170 

"         4'    unequal  arms,  strain  in, 127 

"         "    uniformly  distributed  load, 74 

Leverage,  arm  of, 47 

capacity  of  tubular  girder  by, 369,  370 

graphic  representation, in 

resistance  of  fibre  is  as  the, 223 

Light-well  in  tier  of  floor-beams, 201 

Light-wells,  carriage  beams, 195,  196,  198 

"         "      framing  for,       . 95 

Lignum-vitae,  resistance  of, 120 

Limit  of  elasticity,        . . 212,213,235 

"      "  .      "          deflection  to  the, 237,  243 

"      "         "          in  floor  beam,   . 264 

load  on  beam  at, 246 

"      4'         4t             "      ".  lever  " 248 

strain  beyond  the 313,  315 

"    •"         "          testsofthe, 505 

Limited  application  of  formula  for  value  of  h   in  carriage  beam,  ....  181 


598  INDEX. 

PAGE 

Lines  and  forces  in  proportion 405 

Live  load,  measurement  of  a,  .*.••>.•.     .     ,  •  . 80 

"       "      weight  of  people,     .                     . 84 

Load  and  strain,  various  conditions ....:....  in 

'•     at  limit  of  elasticity  in  a  beam        ...,«,,, 246 

"     "  any  point,  effect  on  beam,     .     ,  • ,     .  56 

"     "    "       "        test  of  rule       ..*.••=• .     ,  57 

"     "    "       "       rule  for  strength,   •-••; 58 

"     "    "       "       safe  rule.     .             ••  •_'.  •.     , 70 

"     "     "       "        strain  at  any  point, .     .     ,     ,  128 

"       to  rupture  a  cast-iron  girder.      .....     T     ...  390 

Load  at  any  point  on  tubular  girder,       .............  371 

'•     "  end  of  rolled-iron  lever, ,               336 

"     "  middle,  pressures, 39 

"     "         "        of  beam .  75 

"     "         " deflection, 242 

"     i4        **    -    M    •  ••      safe  rule, 70 

"     "         "         "     cast-iron  girder 388 

"     "         '*         "     rolled-iron  beam,        .     .     , 331,  332,  333 

"     analyzed,  compound   ..'•.,-.• 178 

44     strains  and  sizes,  compound ,  171 

"     deflection  of  beam  with  distributed 251 

"  lever     "                          255 

"     distributed,  rules  for  size  of  beam  with 253 

safe  rule,       .-  .- 70 

"     equally  distributed,  effect, 58 

"                 "at  middle 60 

"     for  given  deflection  in  a  lever,           247 

"     not  at  middle,  effect  at  middle, 59 

"    **       "         pressure  on  supports e 40,  41 

"     on  beam,  at  middle  and  distributed,       .     .     , 252 

rule  for  distributed , ,  253 

**      "   lever,  distributed  and  concentrated, 255 

"      "    bridge,  Tredgold's 80 

"      "    floor,  components  of 339 

"       Tredgold's  remarks 80 

estimate, 81 

"      "      "       the  greatest  .     .          .- 80 

"       per  superficial  foot 261 

"      "      "       beam,  its  nature .  78 


INDEX.  599 

PAGE 

Load  on  floor-beam,  rule, 78 

"      "    rolled-iron  floor  beam, 340 

"     "    header.      .     . 196 

"     "    carriage  beam,  .                .-..-.,...,..        105,  107 

"     "          "           "       with  one  header, ,  99 

*    roof  per  foot  horizontal,           .     .           ..........  480 

"     inclined  foot  superficial,     ...........  480 

"    supports  arithmetically  computed, ......  456 

"           proportion  of,  ..........      .     .  119 

"          "          from  weight  not  at  middle,     ....;.,....  56 

"    each  support  from  unsyrnmetrical  loading,    .......  451 

•'     per  foot  on  floor,  for  people ,..,..  83 

* "      66  pounds,          ....  87 

4 superficial  of  floor,  70  pounds 88,  264 

on  lever,  promiscuous     ....                ....          ...          .  175 

"     proportioned  to  square  of  depth  of  beam,  ...,.,....  123 

"     upon  a  header,        ....,.,. 96 

"         "     "  bridle  iron,          . 98 

' carriage  beam,               ..........  ^ .  98 

"     "  roof  truss 479,  483 

"         "     each  supported  point  in  a  truss,   .     ,     . 482 

"     tie-beam  of  a  roof.       . ,       481,  483 

Loads  between  the  supports,  dividing  unsyrnmetrical 453 

compared,  concentrated  and  distributed       ..,.,,.,.  61 

Loaded,  framed  girder  irregularly 451 

loo  heavily,  a  beam 243 

Locust,  coefficient  in  rule  for. 261,  265 

resistance  of 120 

Logarithms,  example  of  computation  by, 311 

Mahan's  edition  of  Moseley's  work, 251 

Mahogany,  resistance  of.     .  . 120 

Males,  weight  of     , ,     .     t 83 

Maple,  resistance  o(.       .      . 120 

Mariotte's  theory  of  transverse  strains, 37 

Material,  knowledge  ol  elastic  power  of  any   , 244 

defined,  unit  of, ,     .  29 

Materials  for  important  work  to  be  tested,           244 

"         weights  of  building       .           .     .                504 

Table  XXII..      .......    533,534,535 

of  construction,  weight  of,     .......      78,  So,  339,  340,  379 


600  INDEX. 

PAGE 

Materials  of  construction  of  floors, 502 

"  in  a  roof,  weight  of,         480,  483 

Maximum  moment  defined, 188 

"  ordinate  by  the  calculus, 180,  184 

"  strain  analytically  denned.   ..,..,,',..,.  181.  184 

"  -"       graphically  shown,     ,  .     .     , 178 

compound  load ,  186,  189 

"  "       location  analytically  defined, ,     .    '.     .  179,  184 

"  "       three  concentrated  loads, .     .     202 

"  "       on  middle  one  of  three  headers,  201,  204 

"  "        "  outside  "     "       "         "  196,  204 

"  "of  three  loads  on  carriage  beam, 197,  198 

Maxwell,  reciprocal  frames  and  diagrams  by  Prof 418 

Measure  of  extension  of  fibres, 237 

"         "  forces  in  equilibrium 407 

"         "  symbol  for  safety  tested, 269 

"  resistance  of  cross-bridging, 304 

"         "  strains  in  truss  with  inclined  tie,      .........          475 

"  symbol  for  safety, 239,  269 

Measured  arithmetically,  strains 415 

horizontal  strains     . 412 

Measuring  strains  in  roof  truss,    .     , 485 

Men,  actual  weight  of, , 85 

effect  of  when  marching, 87 

space  required  for  standing  room 82 

Menai  Straits  and  Conway  tubular  bridges 367,  368 

"  "       weight  of  bridge  over, 378 

Merrill's  Iron  Truss  Bridges, 402 

Methods  of  solving  a  problem,  various 72 

Military,  estimate  of  space  required  by, .       83 

"        weight  of,      .     .          ..,,..,.       85 

step,  the  effect  of,       ....,,...     c 86 

Mill  floor,  load  on. .     .     .     , 78,  79 

Minimum  of  strains  in  framed  girder, 426 

area  of  tubular  girder, 383 

Model  of  a  floor  of  seven  beams, 303 

"       iron  tubes  experimented  on, 369 

Modulus  of  elasticity  for  wrought-iron, 232 

"  rupture  by  Prof.  Rankine, 50 

Moment  of  inertia  defined,       314,  319 


INDEX.  60  T 

PAGE 

Moment  of  inertia,  value  of 320 

arithmetically  considered, 314 

geometrical  approximation,   ,     .     , 315 

"         "       "       by  the  calculus, 318,  319 

"         "       "       area  of  parabola ....     322 

"         "       "       shown  graphically, 321 

"        computed, 315,  316,  317 

general  rule  for 324 

"         "       "    .    comparison  of  formulas,          328,  330 

"       proportioned  to  cross-section,    .          ...          .     .          314 

"       "       resistance  to  flexure,  ....     314 

"      "       for  rolled- iron  beams. .       326,  328 

"      "         "       "        "         "        load  at  middle,      ....       331,  333 

"      "         "       "         "         "         Table  of, 498 

41         "      "         "    flange  and  web 327 

"         "      "          "    rolled-iron  header 349 

"         "       weight  defined 47 

"  "         on  lever,      . in 

"         "          ".        arm  of  lever, 56 

"  at  middle  of  beam, 48 

Moments  of  compound  load, 188 

capacity  of  tubular  girder  by 369 

load  at  middle,  tubular  girder  by, 370 

Momentary  extra  strains,      . 87 

Mortising,  the  weakening  effect  of,    .     . 195 

damaging  to  a  header. t 97 

carriage  beams  to  be  avoided -  .     .       98 

Moseley.  moment  of  inertia,  by  Canon 328 

"        modulus  of  rupture  by  Prof. 50 

"        symbol  for  strength    ' 49 

Moseley's  work  on  Mechanics  of  Engineering  and  Architecture,       ,       251,  255 

Movement  of  men,  effect  of .          86 

Negative  equals  adding  a  positive,  deducting  a .          428 

Neutral  axis,  distance  from,  315,  318,  319,  324 

"         "      in  a  framed  girder 402 

"     flange  to  be  distant  from, 313,  314 

line * 45 

"         ''      denned 37 

•'         "      distance  of  fibres  from 222 

"         "      at  middle  of  depth, 45 


602  INDEX. 

PAGE 

Neutral  axis  at  any  depth,  effect 46 

•'         "      in  a  deflected  lever, 236 

"         "         "    tie-beam,  fibres  near        488 

New  England  fir,  experiment  on, 233 

Oak,  coefficient  in  rule  for,        . 261,  265 

"     live,  resistance  of,  ....          120,  121 

Office  buildings,  formula  for  solid  floors  of.         502 

rolled-iron  beams  for, 495,  498 

"     Table  XVIII., 526,  527 

Openings  in  floors,  framing  for,     . ,       95 

Ordinate.  location  of  longest,  compound  load, 182,  184.  185 

Ordinates  measure  strains,     128,  130,  134,  136,  139,  140,  167,  168,  172,  173,  177, 

179,    l8l,    183,    184,    197,   198.   201,   202 

Ordinates  measure  strains  in  lever 175 

Panels  in  a  framed  truss,  number  of, 426,  428 

Parabola,  a  polygonal  figure 161 

"          the  curve  of  equilibrium  is  a 416 

expression  for  the  curve,        160 

form  of  scale  of  strains, 177 

side  of  beam  from  a 124,  163 

"     "  lever      "         . 124.  172,  173 

defines  strains  in  lever, , 169 

form  of  web  of  cast-iron  girder  is  a 392 

Parabolic  curve,  moment  of  inertia,       .          322 

"              "       limits  the  strains, , 184,  197,  201 

"      form  of  floor  arches, 346 

Parallelogram  of  forces  in  framed  girders,      , :  404 

People  as  a  live  load,  weight  of 84 

to  weigh  them, 81 

floors  covered  with 340 

required  for  a  crowd  of, ' .  77 

on  floor,  crowd  of, 81 

their  weight,    .     .     .                81 

per  foot,   ....... 83 

their  weight,  authorities, 82 

Philadelphia,  iron  beams  at  Exposition  at, .     .  313 

Phoenix  Iron  Co.,  beams  tested  by 500 

Planning  a  roof,  general  considerations,     ..... 478 

an  example  in • 481 

Plaster  of  Paris,  fire-proof  quality  of, f     .     .  501 


INDEX.  603 

PAGE 

Plastered  ceiling  of  a  room, 303 

Plastering,  weight  of,       79 

perceptible  deflection  injurious  to,     . 260 

Plate  beam  formed  with  angle  irons, 312 

girder  or  beam,        .                 .                      . 367 

and  tubular  girders  compared.            377 

"    over  a  tubular  girder,  advantages  of  a 377 

Polygon   forces  shown  by  a  closed    . 418 

funicular  or  string 408 

"         lines  in  force  diagram  form  a  closed 485 

Polygonal  figure,  parabola, 161 

Pores  of  wood,  size  of, 120 

Position  of  weight  on  a  beam, 54 

Positive  quantity,  deducting  a  negative  equals  adding 428 

Post,  rule  for  thickness  of  a, 447,  449 

Posts,  Baker's  formula  for, 446 

Precautions  in  regard  to  constants, 243 

Precise  rule  for  carriage  beams,  for  dwellings 273,  282,  285 

"    first  class  stores 275,  282,  286 

with  two  equidistant  headers,      ....  287 

"     headers,  for  dwellings,       .     .     .  292 

"   first-class  stores,      .*  292 

and  one  tail  beam,        .  283 

equidistant     headers    and    one 

tail  beam 290 

"     headers  and  two  tail  beams,  .     .  279 

Pressure,  conditions  in  loaded  beam, 39 

on  support  from  load  not  at  middle 40,  41 

Problem,  various  methods  of  solving  a, 72 

Promiscuous  load,  scale  of  strains, 175 

"     on  lever 167 

Proportion  between  flanges  and  web, 313 

Proportions  of  a  framed  girder, 422 

Quetelet  on  weight  of  people, 83 

Rafters,  dimensions  of, 490,  491 

increased,  compressive  strain  in   ...           .     . 474 

"        to  be  avoided,  transverse  strains  in    .           460 

"         "   "    protected  from  bending, 479 

Rankine  on  compression  of  materials, 446 

"    converging  forces 418 


604  INDEX. 

PAGE 

Rankine  on  modulus  of  rupture 50 

"   moment  of  inertia 328 

Rate  of  deflection  per  foot  lineal 239,261,264,267,342 

11     "           "           in  floors 240 

"     "    rise  in  brick  arch  in  floor 346 

Ratio  of  depth  to  length  in  tubular  girders 382 

Ray's  Algebra  referred  to 476 

Reaction  of  fibres  on  removal  of  force 213,  222 

"         from  points  of  support .     .    58,  465,  466,  470,  473 

"         of  supports  equal  to  load 39 

"          "          "            "       "   shearing  strain 119 

"         "          "         from  unsymmetrical  loading, 451 

"          "          "         of  framed  truss 415 

Reciprocal  figures  explained, 422 

frames  and  diagrams, 4*8 

"           lettering  of  lines  and  angles,    .          4J8 

Resistance  of  materials 53 

"            **          "           to  destructive  energy, 68 

"           its  measure 53 

"           directly  as  the  breadth 33 

"           increases  more  rapidly  than  the  depth, 34 

"  as  the  area  of  cross-section, 31,  32 

"  not  as  the  area  of  cross-section, 31,  33 

"           to  compression,       .     . „  .  45 

"  extension 45 

"          "           and  compression.          229 

"           "             .*•  v                           equal, 45 

"            "          "           or  to  deflection 222 

summed  up,          .     ,           225 

"            "  flexure 235 

"         "        rules  for 242 

"            "         "        value  of    F,    the  symbol  of, .  230 

"         "        of  floor  beams 260 

of  a  lever,  rule  for,      ....      * 48 

to  rupture 266 

elements  of,  . 46 

"   cross-strain  shown 47 

of  fibres  to  change  of  length, 46 

"       "      as  the  depth  of  beam. 43 

"       "      directly  as  the  depth, .     .  36 


INDEX. 


605 


PACK 

Resistance  of  fibres  to  extension  and  compression 35 

"           "       "      "          "          expression  for, 224 

"          to  extension  as  the  number  of  fibres,            222 

"          as  the  distance  of  fibres  from  neutral  line, 222 

"          of  cross  bridging,  principles  of, 304 

"          increased  by  cross  bridging, 305,  310 

of  a  bridged  beam 304 

"          in  cross-bridging,  number  of  beams  giving 309 

Rise  of  brick  arch  in  floor,  rate  of, 346 

Rivet  holes  in  iron  girders,  allowance  for 368 

Rolled-iron  beams,  chapter  on, 312 

"     beam,  history  of  the 313 

"         "     beams  preferable  to  cast-iron, 399 

"         "         4<       means  of  manufacture, 386 

"         "         "       have  superseded  c^st-iron,    ....          386 

"         "         "       to  be  had  in  great  variety, 313 

"    .     "         "       distance  from  centres, ......  342 

"         "     beam,  moment  of  inertia  for, 326,  328 

"         "         "     weight  of, .     .  340 

"         "         "      plate  beam  and  tubular  girder, 367 

"         "         "      load  at  any  point, 333,  334 

"     ' Table  XVII., 335 

"         "          "         "     "  middle 331 

"         "         "         "     distributed,         -  337 

"         "     beams  for  dwellings,  etc 498 

distance  from  centres, 343 

"         "         "         "  first  class  stores 498 

distance  from  centres,      ....  344 

Table  of  elements  of, 498 

"         "         "        elements  of,  Table  XVII., 524,  525 

Tables  of, 495 

•"•      for  dwellings,  Table  XVIII 526,527 

"  first-class  stores,  Table  XIX... 528,  529 

"         "     lever,  load  at  end. 336 

"          "         "         "     distributed 338 

"         "     headers  for  dwellings 349 

"         "           "         "    first-class  stores, 350 

"         "     carriage  beam  with  one  header,  for  dwellings 351 

"         "           "            "       "        "         "         "   first-class  stores,       .     .     .  352 


606  INDEX. 


PAOB 


Rolled-iron  carriage  beam  with  two  headers  and  one  set  of  tail  beams,  for 

dwellings,  etc..  . 358 

"      two  headers  and  one  set  of  tail  beams,  for 

first  class  stores, .     .     359 

"      two  headers  and  two  sets  of  tail  beams,  for 

dwellings,  etc 354^  356 

two  headers  and  two  sets  of  tail  beams,  for 

first  class  stores, 355 

"      three  headers,  for  dwellings,    .     .     .       360,  362 
"    first-class  stores,    .       361,  364 

Roof,  general  considerations  in  planning  a 478 

an  example  in  planning  a .gj 

beams,  increase  in  weight  of 478 

trusses,  chapter  on,     .     .     .v  , 459 

comparison  of  designs  for 450 

selecting  a  design  for 479?  48X 

considered  as  girders 4c(, 

with  and  without  tie-beams 459 

truss,  force  diagram  for  a „ 46i 

horizontal  strain  in 473 

supports,       ....    '.,.    .  : II9 

Rule  for  floor  beams,  using  the g~ 

Rules  for  rupture,  various  conditions Gg 

Rupture  the  base  of  rules  for  strength, 77 

resistance  to,    ... .     .     ,  221,266 

"  theory  of,  , ^7 

"  elements  of, 46 

modulus  of,  by  Prof.  Rankme 5o 

equilibrium  at  point  of,      .     .  kg 

by  compression  and  tension, o     3x3 

"    cross  strain, ZII 

its  resistance,  tension, 4Q 

and  flexure  compared, 211    267 

rules  compared ....       235    203 

compared,  weights  producing, 237 

ensues  from  defective  elasticity 2I2 

beams  of  warehouses  to  resist 260 

resistance  of  carriage  beam  to,  rule  for 100 

of  cast  iron  girder,  load  at  any  point 300 

relation  of  flanges, •      .     .     387 


INDEX.  607 

PAGE 

Safe  load  at  any  point  on  cast  iron  girder .  391 

"    distributed  load,  effect  of  at  any  point  on  cast-iron  girder 391 

"    and  breaking  loads  compared 68 

"    load,  value  of  a,    the  symbol  for  a,     .           235 

"    loads,  rules  for  strength, , 7° 

"    by  rules  for  strength  yet  too  small,  beam 77 

"    beam  should  appear  cs  well  as  be, 235 

"    load  on  tubular  girder 3°9 

Safety  in  floors 28 

precautions  to  ensure 244 

measure  of  symbol  for, ,     .     .     .     .  239 

a,    in  terms  of   B   and    F,    symbol  for, 268 

cautions  in  regard  to  symbol  for 71 

Sagging  of  framed  girder  from  shrinkage, 450 

Scale,  strains  measured  by, m 

of  depths,  compound  load  on  lever, 172 

"  strains  and  the  calculus, 161 

"       "      "       demonstrated, 134 

,"       "      "       applied  practically, .       411,  414 

"       "       "       to  be  carefully  drawn 412 

"       "       "       for  depths, 132 

"       "       "       made  from  given  weights, 409 

"       "        load  at  any  point, 128 

"       "        promiscuous  load,    .     .  167,  175 

"        for  two  weights 133 

"      "        distributed  and  one  concentrated  load, 179 

"  "     two  loads 184,  187 

"       "       compound  load  on  beam, 177 

"      "    lever 172,  174 

"       "       carriage  beam  with  three  headers,   .      .      .  197,  198,  201,  202,  208 

"  weights  for  a  force  diagram 483 

Scientific  American  quoted, 302 

Set  produced  by  strain  on  materials 505 

Shape  of  beam  elliptical, 164 

"  side  of  beam  under  a  distributed  load 162 

"     "      "       from  parabola 163 

"       "     "      "       graphically  shown, 122 

"  lever  a  triangle  under  a  distributed  load,             170 

Shearing  and  transverse  strains 116 

strain  equals  reaction  of  support, 119 


6o8  INDEX. 

PAGE 

Shearing  strain  graphically  shown,    ...          .          115 

"  "     provided  for 123 

•'  "     at  end  of  beam, 118 

•'     .         "      in  tubular  girder 374,  375 

Shrinkage  of  timbers,  derangement  from,  . 450 

Side  of  beam  graphically  shown,  shape  of 122 

"     pressure,  resistance  to, 119 

Size  and  strength,  relation  of, 31 

Skew-back  of  brick  arch  in  floor, 345 

brick  arch  footed  on, 398 

41    '    "      tie  rod  to  hold  arch  on 398 

Slate,  brick  arches  keyed  with 345 

"      on  roof,  weight  of,      ...         „          480 

Sliding  strains,  experiments  on, 505,  506 

"  "         in  Georgia  pine,  locust  and  white  oak,  Table  XXXVIII.,      557 

"  '*          "  spruce,  white  pine  and  hemlock,  "      XXXIX.,      .     558 

"       surface,  breaking  weight  per  square  inch,  "      XLV., .     .     .     564 

Snow  on  roof,  weight  of, 480 

Soldiers  on  a  floor,  weight  of, 83 

Solid  and  laminated  beams  compared, 34 

"     timber  floors  not  so  liable  to  burn 500 

"          "  "      reduction  of  formula  for, 501,  502 

"          "  "      should  be  plastered, 501 

"      Table  for,        .... 504 

"          "  "      thickness  of 500,  501 

.  "      Table  XXI. 532 

Space  on  a  floor  occupied  by  men .     , 85 

"       "  "     "      required  by  people 83 

"    men  when  moving, ,    \  86 

"  "     "    reduced  by  furniture 88 

Spruce,  coefficient  in  rule  for 261,  265 

resistance  of, ,     .     .     .       120,  121 

"         beams,  weight  of, 79 

"        floor  beams  and  headers 495 

Square  timber,  rule  for  strength  of .     .     .       72 

Squares  of  base  and  perpendicular  of  triangle, 487 

Stability  to  be  secured  in  buildings 27,  28 

Stable  and  unstable  equilibrium,  .          416 

Stair  header,  strain  on  carriage  beam, 197 

Stairs,  framing  for, 95 


INDEX.  609 

PAGE 

Stairway  opening  in  floor, 201 

"       openings,  carriage  beams, 195,  196 

Step,  effect  of  military 86 

Stiffness  and  strength  compared,        267,  268 

"          "          resolvable,  rules  for, 270 

"       differ  from  rules  for  strength,  rules  for 235 

"       requisite  in  tloor  beams, • 260 

Stores  for  light  goods  same  as  dwellings, 264 

"      floor  beams  for  first-class 264 

"      headers  for  first  class 271 

"      carriage  beams  for  first-class,  precise  rule  for,       ,  275,  282.  286 

"             "             il       with  one  header,  for  first-class           .     .          ...  273 

"             4i             "          "    two  headers,  "     "         "     precise  rule,        ,     .  292 

equidistant  headers,   for   first-class,    pre- 
cise rule, 289 

"             •'             "           "       "     headers  and  one  tail  beam,  for  first-class.  278 

"       "     headers  and  two  tail  beams;  for  first-class,  276 
"             "                         "    three   headers,    the    greatest    strain    being   at 

middle  header 299 

"  "    three   headers,    the    greatest   strain    being   at 

outside  header, 295 

"       load  on  tubular  girders  for  first-class 380 

"       tie-rods  of  floor  arches  of  first-class 348 

Georgia  pine  beams  for  first  class,  Table  VIII., 515 

"             "           "     headers    "      "      "           "      XVI.,        523 

"       hemlock  beams              "       "       "           "            V.,        ...."..  512 

"         headers            "       "       •'           •*      XIII 520 

"       spruce  beams                 "      ''      l'           "         VII., 514 

headers              "      "      "           •'         XV., 522 

"       white  pine  beams                                      "          VI., 513 

-     headers        "       XIV.,       .     .     ..  '  ,     .     .  521 

"       rolled-iron  beams         "      "      "           "       XIX., 528,  529 

Straight  line,  equation  to  a,  .           r  171 

Strains  useful,  knowledge  of  gradation  of,       .     .     .     . 433 

"         by  movement,  increase  of 84 

"         analytically  defined , 137 

"         arithmetically  computed, 168,  415 

"         graphically  represented,        127 

"         checked  by  computations,  graphical 132 

"        graphically  shown,  shearing 115 


6 10  INDEX. 

PAGE 

Strains  measured  by  ordinates, 128,  130,  167,  168,  177,  201,  202 

"                "           "    lines 405 

proportioned  by  triangles 486 

Strain  analytically  defined,  maximum     .  181,  184 

graphically         "  . ,178 

"      analytically         "        location  of,        .     .     , 179,  184 

at  any  given  point  graphically  shown 127 

"    '      point  from  a  distributed  load, 161 

Strains  demonstrated,  scale  of.                     134 

"       from  given  weights,  to  construct  scale  of, ,  40-) 

promiscuous  load, 167 

distributed  load,  by  the  calculus 157 

two  weights,      .     .     . 101 

"         "          graphic 133 

of  compound  load  analyzed, 178 

••-<•'       "           "               "     maximum.                186,  189 

"     greatest  at  concentrated  load 181 

and  dimensions,  compound  load,    .           171 

Strain  at  first  weight,  with  equal  weights, 147 

.*"       "  second  weight,  with  equal  weights, 148 

"  any                                                                      150 

Strains  in  a  beam,  graphical „ 114 

Strain     "  "       "       loaded  at  any  point.             , 128 

Strains  "  beam  and  lever  compared,       .     .          336 

*'  lever  measured  by  scale ,  m 

"       "      "     computed   "  calculus 169 

"       "  levers,  graphical 167 

"       "  double  lever,  graphical 113 

"       "  lever  with  unequal  arms,    .     . 127 

promiscuously  loaded.  . 175 

"     "      uniformly 168 

like  two  weights  at  ends  of  lever 45 

"      from  three,  headers. 197 

"      in  framed  girder  arithmetically  computed 435 

"        peculiarity  in 432 

tracing  the 437 

"  diagonals  of  framed  girder 436 

"  lower  chord  of  framed  girder 439 

11  upper      "      "       "            "           440 

"  chords  and  diagonals,  gradation  of, 432,  435 


INDEX.  6ll 

PAGE 

Strains  in  roof  truss  compared, 469 

"       "  truss,  arithmetical  computation  of, 486 

"       "  equilibrated  truss 408 

"       "  tie-beam  of  truss,  two 488 

"       "  horizontal  and  inclined  ties  compared, 472 

"       "  truss  with  inclined  tie  may  be  measured, 475 

"       "     "      an  infinite  series, 476 

"       "     "      without  tie  increased,     , 461 

"      from  raising  the  tie  of  truss  increased, .     .  477 

"      in  rafters  increased,  .           460 

"  Georgia  pine,  transverse,  Table  XXIII., 536 

"       "  hemlock.                               Tables  XXXIII.  to  XXXV.,       .     551  to  554 

"       "  locust,                                   Table  XXIV 537,  538 

"       "  spruce,                                   Tables  XXVI.  to  XXVIII.,    .     .     540  to  544 

"       "  white  pine,              "                 "        XXIX.  to  XXXII.,      .   545,  546,  547, 

548,  5-19,  550 

11       "  white  oak,               "            Table  XXV 539 

"  Georgia  pine,  locust  and  white  oak,  tensile,  Table  XXXVI..  .     .  555 

"       "  spruce,  white  pine        "     hemlock,         "             "       XXXVII.,      .  556 

"       "  Georgia  pine,  locust    "     white  oak,  sliding,     "       XXXVIII,    .  557 

"       "  spruce,  white  pine       "    hemlock,         "             "       XXXIX.,  .     .  558 

"       "  Georgia  pine,  locust  "     white  oak.  crushing, "       XL.,      .     .     .  559 

"       "  spruce,  white  pine       "     hemlock,            "          "       XLI  ,    .     .      ..  560 

Straining  beam  in  a  roof  truss.       .     .           460 

"      dimensions  of  a, 490,  491 

Strength,  test  of  specimens  as  to, 29 

as  the  area  of  cross-section 46 

not  as  the  area  of  cross-section, •  .     .     .     .  33 

"         directly  as  the  breadth 33 

increases  more  rapidly  than  the  depth .  34 

"         in  more  common  use    rules  for 235 

and  stiffness  compared.     ....      ........       267,  268 

"         differ  from  those  for  stiffness,  rules  for     . ,     .  235 

"         more  simple  than  those  for  stiffness,  rules  for  .......  235 

"         and  stiffness  resolvable,  rules  for 270 

"    size,  relation  of,            .          ....  31 

"         of  beams,  rule  for 49 

"      "        general  rule  for, 92 

rule  for  load  at  any  point, 58 

"  beam  increased  by  a  device 402 


6l2  INDEX. 

PAGB 

Strength  of  floor,  by  experiment,       , 29 

44          "      "       beams,  rule  for.     ,  • 77 

"          "  beam  and  lever  compared, 55 

14          "  lever,  rule  for,      .     .     ,     .    -.    • 55 

44           "  square  timber,  rule  for 72 

44          "  wood,  unit  of  material , 30 

String  polygon,  funicular  or           408 

Strongest  form  for  a  floor  beam , ^    .  312 

Struts  of  timber  under  pressure ......    V     .".  404 

formula  for  compression  of 447 

"  "          "   thickness  of.       •.•.:.-." *     •     •     -  447.  449 

"       of  framed  girder,  compression  in .     .     .     .     .  445 

"       or  straining  beams  in  trusses,      ... 460 

"      and  ties  form  triangles  in  a  girder 425 

"      in  trusses  prevent  bending  of  rafters,  .          479 

Superficial  foot   load  per      .    ..'"'                       . 88 

44             "         "     on  floors  per 261 

"             "         "     200  pounds  per, 264 

"             "         "    250       "          "       . 264 

41                        "    on  roofs  per  inclined,    .,..., 480 

"             "       weight  of  people,     .          , 82 

44  "        "  tubular  girder  per 378.  379 

Superimposed  load  on  floor,.    . : •'.•  '-; ; '' 78,  80 

44        "  tubular  girder 379 

Superincumbent  load  on  floor 339,  502 

Support  in  a  roof  truss,  points  of, 479 

"         "  "  framed  girder,  points  of, 425 

"        of  a  truss,  weight  upon .     .         464.  466,470 

Supports"  "    ''       division  of  load  upon 461 

unyielding, 119 

reaction  from, 58 

equal  to  load,  reaction  of, 39 

of  framed  truss,      ;t         i:         415 

portions  of  load  on 119 

shearing  strain  equals  reaction  of,    .     .               .     .          ....  119 

Surfaces  of  contact,  resistance  of, 119 

Suspension  bridge  at  Vienna, 82 

44          rod  of  truss,  strains  in.. ,  477 

44             "    4<     "       iron  for 490 

Symbol  of  safety,    a,    the. 267 

41      '*        a,    in  terms  of    B    and    F, 268 


INDEX.  613 

PAGE 

Symbol  of  safety,  value  of    a,    the 69,  235,  239,  269 

'         cautions  in  regard  to, 71 

"       "    unit  of  materials,  the, 49 

Symbols,  assigning  the 101 

compound  load,  assigning  the 187 

for  two  weights,  " 105 

"    headers,  " 108 

"    three      "  "....,....   201,  204,  206 

"    beam  and  lever  compared,     .     .     .          49 

Symbolic  expression,  moment  of  inertia,  .  . 314 

System  of  trussing  a  framed  girder, 425 

Tables,  chapter  on  the,    .• 495 

of  beams  for  dwellings,  etc 495 

"       "      of  wood. , 495 

"        "       "      for  first-class  stores 495 

"         "  rolled-iron  beams 495 

save  time  of  architect,      .  495 

Tables.       .  ....... 507 

Table  I.,  hemlock  beams  for  dwellings ,, 508 

"      II.,  white  pine  beams  for  dwellings 509 

"      III.,  spruce  510 

"      IV..  Georgia  pine  "         "  ........     511 

"      V.    hemlock  "         "    first-class  stores 512 

"      VI  ,  white  pine       "  "       "          " 513 

"      VII.,  spruce  "         "      "       "          " 514 

"      VIII.,  Georgia  pine  beams  for  first-class  stores, 515 

"      IX..  hemlock  headers  for  dwellings. 516 

"      X.,  white  pine       "  517 

"      XL,  spruce  518 

"      XII,,  Georgia  pine  headers  for  dwellings,       ....,..,.     519 

"      XIII.    hemlock  headers  for  first-class  stores 520 

"      XIV..  white  pine     "          "      "        "         '*  ,     . 521 

XV  ,  spruce  " 522 

"      XVI ,  Georgia  pine  headers  for  first-class  stores, 523 

"      XVII.,  elements  of  rolled-iron  beams, .524,525 

"      XVIII.,  rolled-iron  beams  for  dwellings, 526,  527 

*'      XIX.,          "         "         "         "    first  class  stores 528,  529 

"      XX.,  constants  for  use  in  the  rules 530,  531 

"      XXL,    solid  timber  floors, 532 

"      XXIL,  weights  of  building  materials, 533.534,535 


614 


INDEX. 


Table  XXIII.,  transverse  strains  in  Georgia  pine, 

"  XXIV.. 

-.;;••  XXV., 

; .-«'  XXVI.,         •*. 

r-»*  XX  VII., 

<«•••_  XXVIIL,     " 

c  v".  XXIX, 

>«  XXX., 

-  "  XXXI., 

.."  XXXII., 

.;:"  XXXIIL,     " 

-.-V  XXXIV,      " 

,-V  XXXV.,      .  " 

-•!•••  XXXVI.,  tensile 

;"  XXXVII.,      " 

"  XXXVIII..  sliding      '# 

"  XXXIX.,         "    . 

"  XL.,  crushing  " 


PAGE 

•         • 536 

locust 537,  538 

white  oak 539 

spruce 540 

•     •     . 541,  542 

543,  544 

white  pine, 545 

546,  547 

.     •  548.  549 

"       " 550 

hemlock,     . 551 

552.  553 

•     •  •  554 

Georgia  pine,  locust  and  white  oak,  .  555 

spruce,  white  pine  and  hemlock.    .  .  556 

Georgia  pine,  locust  and  white  oak,  .  557 

spruce,  white  pine  and  hemlock,    .  ,  558 

Georgia  pine,  locust  and  white  oak,  .  559 

spruce,  white  pine  and  hemlock,  .  560 


XLI., 

\  '*      XLIL, transverse  breaking  weight  per  unit  of  material.    .     .     .     .     .  561 

XLIIL,  deflection,  values  of  constant,    F. 562 

XLIV.,  tensile  breaking  weight  per  square  inch  area, 563 

XLV..  sliding         "               "         '     surface,      ....  564 

XLVL.  crushing  weight  per  square  inch  sectional  area 565 

Tail  beams,  definition  of,      . 95 

two  headers  and  one  set  of, 106 

"    sets  of, 104 

three  headers  and  two  sets  of, 200 

Tangent  defined,  point  of  contact  with 179,  184,  198 

Tensile  and  compressive  strains 408 

strains,  experiments  on, -505 

in  Georgia  pine,  locust  and  white  oak,  Table  XXXVI.,  .  555 

"         "   spruce,  white  pine  and  hemlock,  Table  XXXVII.,  .     .  556 

'-•"  -    breaking  weight  per  square  inch  area,  Table  XLIV., 563 

strength  of  wrought-iron 347 

Tension  measures  rupture 49 

"         graphically  shown, ....  115 

in  a  framed  girder,  resistance  to 443,  445 

dimensions  of  parts  subject  to 487 

"         and  compression,  rupture  by,        313 


INDEX.  6l5 

PAGE 

Tension  and  compression  in  cast-iron, 387 

in  wrought-iron, 117 

"  tubular  iron  girder 37° 

"  bottom  flange  or  tie-rod 39°.  397 

rule  for  shearing  based  on 117 

Testing  machine 5°4 

Test  of  deflection  in  a  lever 245 

Tests  of  the  materials  used  desirable 69 

"      should  be  made  for  any  special  work, 5°° 

"of  value  of  symbol  of  safety,    ....          69,  269 

Three  headers,  the  greatest  strain  at  middle  one 201,  204,  207 

"  "       "  outside     " 196,  204 

"       equal  weights  on  beam 141 

weights,  graphic  strains  from 138 

Ties  compared,  strains  in  horizontal  and  inclined 472 

"     in  trusses  extended  through  to  rafters 460 

Tie-beam,  importance  of  a, 404 

"        "      tensile  and  transverse  strains  in  a,                      488 

''        '•      of  roof,  load  upon  the, 48-1.  483 

"        <l       "     "      truss,  strains  in 488 

"        "      "    truss,  built  up      .     .          489 

"        "       "       "       manner  of  building, 489 

Tie-rod,  effect  of  elevating  the 477 

"     "     of  brick  arch,  area  of  cross  section, 347 

«     «'      ««       "        "      where  to  place, 348 

"     "      "       "         "       in  floor 346 

11     "      "       "        !<      on  skew-back 398 

"     cast-iron  arched  girder  with 396 

"     "     of  iron  arched  girder 39°.  397 

Timber,  experiments  on.      ......           30 

fl9ors,  thickness  of  solid,     .          500 

Transverse  breaking  weights  per  unit  of  material,  Table  XLII.,   ....  561 

force,  test, 29 

strain,  the  philosophy  of, 312 

"         experiments  by, 504 

"               "         resistance  to, 47 

"         by  rupture  or  deflection, 211 

"               "         lever  principle, 38 

"         in  a  tie-beam 488 

"    cast-iron, 387 


6l6  INDEX. 

PAGE 

Transverse  strains,  the  object  of  this  work 28 

"  in  framed  girders, 402 

"  "          shearing  and 116,  118 

in  Georgia  pine,  Table  XXIII 536 

"  locust,  Table  XXIV 537,  538 

"  *'          "  white  oak.  Table  XXV 539 

«•  spruce,  Tables  XXVI.  to  XXVIII 54010544 

"  white  pine..  Tables  XXIX.  to  XXXII.,  545,  546,  547,  548, 

549.  550 
"  hemlock,  Tables  XXXIII.  to  XXXV.,    .     .     551  to  554 

"  strength  of  beams,  rule  for, 48 

Tredgold  an  authority  on  construction, 81 

"         on  compression  of  materials, 446 

"         "    cast-iron. 386,  387.  396 

"         has  written  on  roofs .          459 

Tredgold's  "Carpentry." ...          417 

"          estimate  of  load  on  floor  81 

"          rate  of  deflection.    .     . 240 

"  remarks  on  load  on  floor. 80 

"  value  of  elasticity  of  iron. ,     .          232 

Trenton  Iron  Works,  beams  tested  by.       .     , 500 

Triangle,  measure  of  extension, 221 

"         showing  elongation  of  fibres 236 

"         resistance  of  fibres  as  area  of,     .......          .     .       222,  223 

"         distributed  load,  side  of  lever  a, 170 

"         of  forces  in  framed  girders, 404 

Triangles  "       "      Professor  Rankine,        418 

"         "    strains,    ...          128,  168,  413 

"         in  proportion  to  strains 486 

"         "  framed  girder,  series  of,     ....          425 

Trimmers,  definition  of,  .     . , 96 

"  and  headers, 266 

"     bridle  irons, 98 

Truss,  arithmetical  computation  of  strains  in  a 486 

"       graphically  shown,  forces  in  a, 417 

"       should  have  solid  bearings  for  support, 478 

"       with  inclined  tie,  vertical  strain  in 474 

"       without  tie,  increased  strains  in,      .     .' •     .     461 

"       load  upon  each  supported  point  in  a, 482 

"         "         "       "       support  of  a, 464,466,470 


INDEX.  617 

PAGE 

Trusses  for  roofs, 459 

"        load  upon  roof 479.  483 

"        points  of  support  in  roof 479 

"        division  of  load  upon  supports  of  roof 461 

"        distance  apart  for  placing  roof 478 

"        required,  number  of  roof      ..,,,.          481 

dimensions  of  rafters,  braces,  etc.,  in 490,  491 

Trussing  a  framed  girder,  ,     .     .     .       417,  425 

Tubular  iron  girder,  history  of, ,     .     .     .     .     367 

44  "         "         useful  for  floors  of  large  halls,    .          367 

"  "         **         coefficient  of  strength,  ....      . 368 

"  "         "         constants J  ........     369 

"  "         '*         by  leverage,      .     .          ....     369 

"  "         "         '*    principle  of  moments, 369 

"  *'         "         "    moments,  load  at  middle,  .     .     , 370 

"  "         "         allowance  for  rivet  holes 368 

«*  "         "         shearing  strain  in 374,  375 

'*         buckling  of  sides  of, 377 

"         "         uprights  of  T  iron  in  sides  of, 377 

"         "         economical  depth ,     « 382 

"         ratio  of  depth  to  length 382 

"    '     '*         approximated,  weight  of.  .          378 

"         "         per  superficial  foot,  weight  of, 378,  379 

"         "         minimum  area  of .     .          .     .       382,  383 

"         "         tension  in  lower  Mange  of. 370 

"         "         top  and  bottom  flanges  equal, 371 

"         "         construction  of  flanges  of, 374 

"         "         thickness  of  flanges  of, 373 

**         "         strain  in  web  of. 374 

"         "         construction  of  web  of 376 

"         "         thickness  of  web  of, 375 

"         "  "         "   plates  of, ,     .     382 

"        "        load  at  middle, 367 

44      "       "        common  rule, 368 

"      "  any  point 368,  371 

"     uniformly  distributed, 368,  372 

"         "  size  at  any  point,     .     .     .     372 

**  "         "        weight  of  load  on  floor, 377 

**  "         *'         for  floors,  rule  for, 377 

*'  "         "          "   dwellings,  etc.,  load  on, 380 


6l8  INDEX. 

PAGK 

Tubular  iron  girder,  for  dwellings,  etc.,  rule  for, 380 

"           "         "         compared  with  plate  girder,  . 377 

advantages  of  plate  girder  over, 377 

•'           "     girders,  chapter  on 367 

Two  loads,  graphic  strains,       .                     133 

Two  weights,  their  effect  at  location  of  one  of  them, 103 

Union  Iron  Co.  of  Buffalo,  large  beams,    ,  '  .  '  , 313 

Unit  of  material,  symbol  of,     ...               49 

"     "         "           Moseley's  symbol  of .  49 

"     "         v'           index  of  strength,            .     .           50 

41     "         "•          measure  of  strength, 30 

strength  and  size  of, 46 

"     "•         "           size  arbitrary, 32 

dimensions  adopted 53 

and  weight 53 

harmony  necessary  between  piece  taken  and     ....  54 

breaking  load  of 69 

resistance  of, 53 

Unknown  quantities,  eliminating, 90 

Unstable  and  stable  equilibrium, 416 

Unsymmetrical  loading,  reaction  of  supports  from, 451 

Unsymmetrically  loaded  girder,  force  diagram  of, .     .  455 

Unyielding  supports  for  load 119 

Value  of    h    limited  to  certain  cases 181 

Versed  sine  of  brick  arch 346 

Vertical  effect  of  an  infinite  series, 476 

strain  in  truss  with  inclined  tie 474 

Vienna  suspension  bridge, , 82 

Von  Mitis,  testimony  as  to  load  on  bridges, 82 

Wade's  experiments,  Major 500 

Wales,  tubular  bridges  of, 367,  368 

Walker,  James,  testimony  on  loading 82 

Walls,  bearing  surface  of  beams  on 121 

of  building  raised  to  permit  horizontal  tie, 478 

pushed  out  for  want  of  tie,        404 

Walnut,  resistance  of, 120 

Warehouse  beams  to  resist  rupture, 260 

load  on  floors  of 78,  79,  339 

Web,  moment  of  inertia  for  the 327 

"      and  flanges,  proportion  between 313 


INDEX.  619 

PAGE 

Web  to  connect  flanges  and  resist  shearing, 314 

"      usually  larger  than  needed 314 

and  flanges  in  cast-iron,  relation  of, 387 

of  cast-iron  girder,  form  of, 392 

"       "  tubular  girder,  strain  in 374 

construction  of, 376 

thickness  of, 375 

"       "  "       area  of, 382,  383 

Weight  and  depth  of  beam,  relation, 36 

"        in  proportion  to  depth 44 

"        its  effect  and  position 53 

"       "      at  end  of  lever 58 

"        on  a  lever,  rule  for 256 

"         "   "  beam,  its  position, 54 

not  at  middle,  effect  on  supports, 56 

"        at  middle  of  rolled-iron  beam, 331,  332 

on  each  support  of  a  truss, 464,  466,  470 

"        of  military 85 

"         "  people,  live  load, 84 

"  beams  per  superficial  foot, 79 

"        "  floors  in  dwellings,        80 

"        "  timber  in  solid  floors,  .  502 

"        "  materials  of  construction  in  a  roof, 480,  483 

"        "  rolled  iron  beams, 339,  340 

Weights  of  building  materials, 504 

"         assigning  the  symbols  for,        101,  105 

"         and  deflections  in  proportion, 304 

"     lengths,  relation  between, 65,  66,  67 

pressures,  equilibrium, 38 

"     strains  put  in  proportion, •     .     .     .  409 

in  proportion  as  arms  of  levers, 39 

at  location  of  one,  effect  of  two 101 

"         for  force  diagram,  line  of, 464,  466 

producing  rupture  and  flexure  compared 237 

of  building  materials.  Table  XXII., 533,  534,  535 

Whalebone  largely  elastic 211 

White  pine,  coefficient  in  rule  for 261,  265 

resistance  of, 121 

"      experiment  on  units  of, 32 

beams,  weight  of, 79 


620  INDEX. 

PAGE 

White  pine  floor  beams  and  headers.      . 495 

Whitewood,  resistance  of, 120,  121 

Wind,  its  effect  on  a  roof. 480 

stronger  on  elevated  places 481 

Wood,  iron  a  substitute  for, 312 

Woods,  experiments  on  American 504 

Wooden  beams  liable  to  destruction  by  fire, 312 

44      weight  of. 79 

"         floors,  when  solid  not  so  liable  to  burn, 500 

Work  accomplished  in  deflecting  a  lever,        229 

44      to  be  tested,  materials  in  important 244 

Wrought-iron,  modulus  of  elasticity  of, 232 

"       tension, 116 

44      tensile  strength  of, 347 

44       for  ties  in  framed  girders, 443 

Yarmouth,  fall  of  bridge  at, 82 


ANSWERS  TO  QUESTIONS. 


37. — Transverse. 

38. — In  proportion  directly  as  the  breadth. 

39. — In  proportion  directly  as  the  square  of  the  depth. 

4-0. — The  elements  are  the  strength  of  the  unit  of  mate- 
rial, the  area  of  cross-section  and  the  depth. 

The  expression  is  R  =  Bbd'. 

41. — The  amount  is  equal  to  the  total  load. 

42 One  half. 

43. — The  sum  is  equal  to  the  total  load. 

44. — The  portion  of  the  weight  borne  at  either  point  is 
equal  to  the  product  of  the  weight  into  its  distance  from  the 
other  support,  divided  by  the  length  between  the  two  sup- 
ports. 

45.—  R  =  W~ 

*e.-p  =  w^ 

47. — 10000  pounds. 
5000  pounds. 

48. — The  moment,  or  the  product  of  half  the  weight  into 
half  the  length  of  the  beam. 


622  ANSWERS   TO   QUESTIONS. 

49 —  Wl  =  Bbd*. 
50. — 22666!    pounds. 

51. — As   many  times  as  the   breadth  is  contained   in   the 
depth, 

62.— 22781^    pounds. 

63. — 40500    pounds. 

64. — 20250    pounds. 

65. —  5062^    pounds. 

84. — Depth,  6-6  inches;    breadth,  3-3    inches. 

85. — 5-24    inches    square. 

86. — 6-93    inches. 

87. — 4    inches. 

126. — 2    feet     lof    inches. 

127. — 2    feet    4^    inches. 

128.  — 2    feet    ;|    inches. 

129. — 2    feet     if    inches. 

130.— i     foot    8f    inches. 

131. —  i     foot     u|-    inches. 

132. — 2    feet    o    inches. 

133.— i     foot    7f    inches. 

134. —  i     foot    9!    inches. 

135.— 3    feet     i     inch. 

160. — Breadth,    6-78    inches;  depth.    12-34    inches. 

161. — 2-94    inches. 


ANSWERS   TO    QUESTIONS.  623 

162. — 2-86    inches. 
163. — 0-291!    inches. 
164. — 1-763    inches. 
165. — 1-244    inches. 
166. — 3- 1  ii     inches. 
179. — 36720    foot-pounds. 
180. — 36-72     inches. 

(81. —  Ordinates.  Strains. 

For  5  ft.,  18-36  18360  foot-pounds. 

"  6  "  22-032  22032  "  " 

"  7  "  25-704  25704  " 

"  8  "  29-376  29376 

"  9  '"  33-048  33048  " 

182. — 10-733,     10-182    and    9-6    inches    respectively. 
183. — 6-245,    8-062,    9-539    and     10-198    inches    respec- 
tively. 

184. — Depth,    8-1565    inches. 

Weight,    652-218    pounds. 

Shearing-  strain  at  wall,    733  •  783    pounds. 

"  "       "  5  ft.  from  wall,  699-798  pounds. 

185. — 302-222    pounds. 

186. — Shearing   strain,  4973!    pounds. 
Height,    o«93i    inches. 

187. — 1-46    inches. 

204. — io666|    pounds. 

205. — 2666f,     5333^    and     8000    pounds. 

206. — Strain  at    A,  17142!;     at    /?,    22285^. 


624  ANSWERS   TO   QUESTIONS. 

207. — 8571!,    7428f    and    21000    pounds    respectively. 

208. — 12750,  22750,   19000,  8500,  9500,  20500  and    21500 
pounds  respectively. 

209. — 3920  pounds. 

217. — A  parabolic  curve. 

218.— 800,     1050    and     1200    pounds. 

219. — 5  •  1087    inches. 

220.— Elliptical 

228. — o,     300,     600,     900     and     1700    pounds. 
At    the    wall    2500    pounds. 

229. — A  parabolic  curve. 

230. — 1250    pounds. 

231. — Triangular. 

232. — 3450,    6250    and    9450    pounds. 

233. — 200,    1600,    6150    and     11050    pounds. 

269. — 19200       pounds;      located     at     the     concentrated 
weight. 

270. — 7-01     inches. 

287. — Resistance  to  flexure. 

288. — To  any  amount  within  the  limits  of  elasticity. 

289. — The  extensions  are  directly  as  the  forces. 

290. — The  deflections  are  directly  as  the  extensions. 

291. — The  deflections  are  as  the  weights  into  the  cube  of 
the  lengths. 

307. — By  the  power  of  reaction. 


ANSWERS    TO    QUESTIONS.  625 

308. — To  the  number  of  fibres,  to  the  distance  they  are 
extended,  and  to  the  leverage  with  which  they  act. 

309 Wls  =  Fbd'd. 

3(0. — 0-266    of  an  inch. 

315. — The  rules  for  strength  are  the  more  simple. 


3(9.—  r-- 

a 

354.-Formulas  (122.),  (124.),  (125.),  (126.)  and  (120.). 

355.— Formulas  (123.),  (127.),  (128.),  (129.)  and  (121.). 

356.— Formulas  (131.),  (132.),  (133.),  (134.)  and  (135.). 

357.— Formulas  (136.),  (137.),  (138.),  (139.)  and  (140.). 

437.— 12-345     inches. 

438. — 4-176    inches. 

439. — 7-700     inches. 

440. — 8-417     inches. 

441. — 10-779     inches. 

442.— 9-530     inches. 

537.—  -fabd*    (form.  205.). 

538.—  •&(bd'-bld!r)    (form.  213.). 

539. — The  Buffalo    I2j    inch     180    pound  beam. 


626  ANSWERS    TO    QUESTIONS. 

540. — 9475-58    pounds. 

541. — 2004-52    pounds. 

542. — The  Pottsville     io|     inch     90     pound  beam. 

543. — Two     12     inch     170     pound  beams. 

544. — It  should  be  a     15     inch     200     pound  beam. 

545. — A  Paterson  or  Trenton    io£   inch    135    pound  beam. 

574. — 41^    inches. 

575. — 35     inches. 

576,, — At    5    feet  from  the  end  of  girder,    15     inches  each  ; 

"  10      "       "        '•       "     "•       "          26f      "  " 

..     I5  M  U  »  .*         «  M  35 

"   20         "          "  "         "       "          "  40  <;  " 

"  25      "     or  at  middle,  4if      <;  " 

577. — At  end  of  girder,  0-38     inch; 

5    feet  from  end  of  girder,    0-30       " 

-     15       «       «         -     «       «  0-15       " 

"       20        "         "  "       "          "  0-08 

"     25       "     or  at  middle,  o-o         " 

578. — At  5  feet  from  end  of  girder,      8-95    inches; 

"  10      "                 "     "        "  15-34 

"  15      "        "         "     "        "  19.18 

"  20      "      or  at  middle,  20-46 

579.— 4-2155    feet. 

597. — Bottom  flange,     16    X2-I95    inches; 
Top  5^x1-646 

Web,  1-372         u        thick. 


ANSWERS    TO    QUESTIONS.  627 

598.  —  Bottom  flange,     16    x  1-646    inches;. 

Top  Six  1-234 

Web,  1-029  thick. 

599.  —  Bottom  flange,     16    x2«49      inches; 

Top  "  Six  1-867 

Web,  I-556         "         thick. 

600.  —  32-99    inches. 

601.  -^-At  the  location  of  the    25000    pounds  ; 

The  bottom  flange,  16    x  1-663  inches; 

"    top  si  x  i  -  247      « 

"    web,  J-039      "         thick. 

At  the  location  of  the    30000    pounds  ; 
The  bottom  flange,  16    x  1-588    inches; 

"    top  "          Six  1-191 

"    web,  0-992         "          thick. 

602.  —  3-68    inches. 

651.  —  The  strain  in    AB    is    3550    pounds; 
"      "      BC    "   10280 


"      "     AF    "   15240 
««      "  u   10130 


652.—  7-8125    feet. 
653.—  Six. 

654-.  —  The  strain  in    DE    is      3600    pounds  ; 
"      CD    "       5425 


" 


"     14500 

«      2I72, 

At/  "    15700 


628  ANSWERS   TO    QUESTIONS. 

The  strain  in    £S    is     41800    pounds; 
"       "     BL     "     26125 
"    DM    "     39200 

655.— The  strain  in    DE  is  3614    pounds; 

"     CD  "  5420 

"     BC  "  12647 

"     AB  "  14454 

"    AT-4  "  21681 

"    AU  "  15655 

"   "  "  CT  «  35223 

"  ES  "  41746 

26091    " 

39r37 

656.— The  area  at   ^  f/  should   be   16-103  inches; 

36-180 
42-872 

The  size  of   BL  «      6  -  626  x  7  -  95 1  inches ; 

"       "      "    DM       "          «      7-715x9.258      « 

3-004x3-605  " 
4-612x5.534  <{ 
5.651x6-781  " 

The  area  of    CZ>       "          «      0-603  inches; 
"       "      "     AB        "          "      1-611 

688. — The  computed  strain  in    ^(^    is  22535    pounds; 

"     BH    "  18028 

"     ^4^     "  4507 

"     AF    «  18750 

"  "  "        "     ^r    u  15000         " 

"     measured     "        "    AG    "  22500         ' 

"  18000         * 

"  4500 

"  18750 

"  15000 


ANSWERS   TO    QUESTIONS.  629 

689.— The  strain  in    AN   is  44200    pounds  ; 

"     BO     "  38720 

<t  29160         u 
"       5400 

CD     "  13300 

^Jfcf    "  36760 

6Z     "  32240 

BC     "  loooo 

DE     "  26320 


690. — The  strain  in    AJ    is  41000    pounds; 

"     BK    "  36150 

"         u       "     AB     "       4950 

"     AH    "  32400 

"     BC     4<  24700 

"  14500        " 


691. — The  strain  is     16000    pounds. 


692 — Six. 

That  shown  in  Fig.  115. 

The  strain  in    AO    is  96200    pounds: 

«      Bp    «  88200         " 

"     C/7    "  53000 

"     CQ    "  25700 

44     DR    "  17600 

"        "       "     ^^?     "  8000 

"       "     CD     "  8000 

"  81600        " 

"  74800 

DE    "  10200        " 

"  24700        " 


630 


ANSWERS   TO    QUESTIONS. 


AO  should   be    g 

x    15-80) 

BP 

"      9 

f 

x    14-49) 

9x16  tapered  to  9  x 

14 

CH       « 

44     9 

•  13-41, 

or    9x14 

CQ        « 

"     6 

x     8-61) 

DR 

"     6 

y     5-89) 

6x9    tapered  to  6  x 

6 

AB        " 

"      4 

x     6-41, 

or      4x    7 

CD 

"•    4 

6-41 

"       4x    7 

AN 

"'      8.K 

. 

O     O 

'3  x  ii'  77 

9  x  12 

HM       « 

"    U-8 

H    X    IQ.SO 

"           \  £    •/    ">r\ 

1-133    area,  or    ij    diameter. 

2-744     4<        "         2 


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